Use Of Complex Numbers in Binomial Theorem
Writing the binomial expression of \((\cos \theta+i \sin \theta)^n\) and equating the real part to \(\cos n \theta\) and the imaginary part to \(\sin n \theta\), we get
Real part (terms with even powers of \(i\) ):
\(
\cos n \theta=\sum_{k=0,2,4, \ldots}^n(-1)^{k / 2}\binom{n}{k} \cos ^{n-k} \theta \sin ^k \theta
\)
Imaginary part (terms with odd powers of \(i\) ):
\(
\sin n \theta=\sum_{k=1,3,5, \ldots}^n(-1)^{(k-1) / 2}\binom{n}{k} \cos ^{n-k} \theta \sin ^k \theta
\)
\(
\begin{array}{r}
\cos n \theta=\cos ^n \theta-{ }^n C_2 \cos ^{n-2} \theta \sin ^2 \theta+{ }^n C_4 \cos ^{n-4} \theta \\
\sin ^4 \theta+\cdots
\end{array}
\)
\(
\begin{aligned}
\sin n \theta={ }^n C_1 \cos ^{n-1} \theta \sin \theta-{ }^n C_3 & \cos ^{n-3} \theta \sin ^3 \theta \\
& +{ }^n C_5 \cos ^{n-5} \theta \sin ^5 \theta+\cdots
\end{aligned}
\)
\(
\tan n \theta=\frac{{ }^n C_1 \tan \theta-{ }^n C_3 \tan ^3 \theta+{ }^n C_5 \tan ^5 \theta-{ }^n C_7 \tan ^7 \theta+\cdots}{1-{ }^n C_2 \tan ^2 \theta+{ }^n C_4 \tan ^4 \theta-{ }^n C_6 \tan ^6 \theta+\cdots}
\)
We get a very interesting results if in binomial expansion, any variable is replaced with \(i=\sqrt{-1}\) or \(\omega\) (cube roots of unity).
Example 1: Find the sum \(C_0-C_2+C_4-C_6+\cdots\) where
\(\boldsymbol{C}_{\boldsymbol{r}}={ }^{\boldsymbol{\prime} \boldsymbol{\prime}} \boldsymbol{C}_{\boldsymbol{r}}\).
Solution: \((1+x)^n=C_0+C_1 x+C_2 x^2+\cdots+C_n x^n\)
Substitute \(x=i(i=\sqrt{-1})\)
\(
(1+i)^n=C_0+C_1 i+C_2 i^2+C_3 i^3+C_4 i^4+\cdots
\)
Real part: \(C_0-C_2+C_4-C_6+\cdots\)
Imaginary part: \(C_1-C_3+C_5-\cdots\)
\(
(1+i)^n=\underbrace{C_0-C_2+C_4-\cdots}_{\text {Real part }}+i \underbrace{\left(C_1-C_3+\cdots\right)}_{\text {Imaginary part }}
\)
Express in polar form
\(
1+i=\sqrt{2}\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)
\)
Raise to the power \(n\) using De Moivre’s theorem:
\(
(1+i)^n=(\sqrt{2})^n\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^n=2^{n / 2}\left(\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right)
\)
Real part:
\(
C_0-C_2+C_4-\cdots=\operatorname{Re}\left[(1+i)^n\right]=2^{n / 2} \cos \frac{n \pi}{4}
\)
Imaginary part:
\(
C_1-C_3+C_5-\cdots=\operatorname{Im}\left[(1+i)^n\right]=2^{n / 2} \sin \frac{n \pi}{4}
\)
Example 2: Prove that
\(
{ }^n C_0+{ }^n C_3+{ }^n C_6+\cdots=\frac{1}{3}\left(2^n+2 \cos \frac{n \pi}{3}\right)
\)
Solution: Step 1: Consider the cube roots of unity
Let \(\omega\) be a primitive cube root of unity:
\(
\omega=e^{2 \pi i / 3}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, \quad \omega^3=1, \quad 1+\omega+\omega^2=0
\)
Step 2: Use the binomial expansion
\(
(1+x)^n=\sum_{r=0}^n C_r x^r
\)
Step 3: Use roots of unity filter
To pick out the terms where \(r\) is a multiple of \(3(r=0,3,6, \ldots)\) :
\(
C_0+C_3+C_6+\cdots=\frac{(1+1)^n+(1+\omega)^n+\left(1+\omega^2\right)^n}{3}
\)
This formula works because of the roots of unity filter.
Step 4: Compute each term
(i) \((1+1)^n=2^n\)
(ii) \((1+\omega)^n\) and \(\left(1+\omega^2\right)^n\) :
\(
1+\omega=e^{i \pi / 3} \quad \text { (in polar form) }
\)
Check:
\(
1+\omega=1+\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)=\frac{1}{2}+i \frac{\sqrt{3}}{2}=e^{i \pi / 3}
\)
Similarly:
\(
1+\omega^2=\frac{1}{2}-i \frac{\sqrt{3}}{2}=e^{-i \pi / 3}
\)
Step 5: Raise to the power \(n\)
\(
\begin{aligned}
(1+\omega)^n & =e^{i n \pi / 3}=\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3} \\
\left(1+\omega^2\right)^n & =e^{-i n \pi / 3}=\cos \frac{n \pi}{3}-i \sin \frac{n \pi}{3}
\end{aligned}
\)
Step 6: Add them up
\(
\begin{gathered}
(1+\omega)^n+\left(1+\omega^2\right)^n=2 \cos \frac{n \pi}{3} \\
(1+1)^n+(1+\omega)^n+\left(1+\omega^2\right)^n=2^n+2 \cos \frac{n \pi}{3}
\end{gathered}
\)
Step 7: Divide by 3
\(
C_0+C_3+C_6+\cdots=\frac{1}{3}\left(2^n+2 \cos \frac{n \pi}{3}\right)
\)
Sum of Series
Example 3: Find the sum \(C_0+3 C_1+3^2 C_2+\cdots+3^n C_n\).
Solution: The sum looks like a binomial expansion with a factor:
\(
S=\sum_{r=0}^n\binom{n}{r} 3^r \cdot 1^{n-r}
\)
The binomial theorem says:
\(
(a+b)^n=\sum_{r=0}^n\binom{n}{r} a^{n-r} b^r
\)
Here, \(a=1, b=3\). So:
\(
\sum_{r=0}^n\binom{n}{r} 1^{n-r} 3^r=(1+3)^n=4^n
\)
Example 4: If \((1+x)^n=\sum_{r=0}^n C_r x^r\), then prove that \(C_1 +2 C_2+3 C_3+\cdots+n C_n=n 2^{n-1}\).
Solution: Start with the binomial expansion
\(
(1+x)^n=\sum_{r=0}^n C_r x^r=C_0+C_1 x+C_2 x^2+\cdots+C_n x^n
\)
Differentiate both sides w.r.t \(x\)
\(
\frac{d}{d x}(1+x)^n=n(1+x)^{n-1}
\)
and
\(
\frac{d}{d x} \sum_{r=0}^n C_r x^r=C_1+2 C_2 x+3 C_3 x^2+\cdots+n C_n x^{n-1}
\)
Substitute \(x=1\)
\(
C_1+2 C_2+3 C_3+\cdots+n C_n=n(1+1)^{n-1}=n \cdot 2^{n-1}
\)
Example 5: Find the sum \(1 C_0+2 C_1+3 C_2 +\cdots+(n+1) C_n\) where \(C_r={ }^n C_r\).
Solution: Split the sum, Notice that:
\(
(k+1) C_k=C_k+\left(k \cdot C_k\right)
\)
So we can write:
\(
S=\sum_{k=0}^n(k+1) C_k=\sum_{k=0}^n C_k+\sum_{k=0}^n k C_k
\)
Sum of all coefficients:
\(
\sum_{k=0}^n C_k=2^n
\)
Sum of \(k C_k\) :
\(
\sum_{k=0}^n k C_k=C_1+2 C_2+3 C_3+\cdots+n C_n=n 2^{n-1}
\)
\(
\begin{aligned}
&\text { Combine the sums }\\
&S=2^n+n 2^{n-1}=2^{n-1}(2+n)=(n+2) 2^{n-1}
\end{aligned}
\)
Example 6: Find the sum \(1 \times 2 \times C_1+2 \times 3 C_2+\cdots+ n(n+1) C_n\), where \(C_r={ }^n C_r\).
Solution: Step 1: Express in terms of \(k(k+1) C_k\)
\(
S=\sum_{k=1}^n k(k+1) C_k
\)
Step 2: Split \(k(k+1)\) as \(k^2+k\)
\(
S=\sum_{k=1}^n k^2 C_k+\sum_{k=1}^n k C_k
\)
We already know the second sum:
\(
\sum_{k=1}^n k C_k=n 2^{n-1}
\)
Step 3: Find \(\sum k^2 C_k\) using derivative trick
Consider the binomial expansion:
\(
(1+x)^n=\sum_{k=0}^n C_k x^k
\)
First derivative:
\(
\frac{d}{d x}(1+x)^n=n(1+x)^{n-1}=\sum_{k=1}^n k C_k x^{k-1}
\)
Multiply both sides by \(x\) :
\(
\sum_{k=1}^n k C_k x^k=n x(1+x)^{n-1}
\)
Differentiate again:
\(
\frac{d}{d x}\left(\sum_{k=1}^n k C_k x^k\right)=\sum_{k=1}^n k^2 C_k x^{k-1}=\frac{d}{d x}\left(n x(1+x)^{n-1}\right)
\)
Compute the derivative on the right:
\(
\frac{d}{d x}\left(n x(1+x)^{n-1}\right)=n(1+x)^{n-1}+n x(n-1)(1+x)^{n-2}=n(1+x)^{n-2}[(1+x)+x(n-1)]=n(1+x)^{n-2}(1+n x)
\)
So:
\(
\sum_{k=1}^n k^2 C_k x^{k-1}=n(1+x)^{n-2}(1+n x)
\)
Step 4: Substitute \(x=1\)
\(
\sum_{k=1}^n k^2 C_k=n(1+1)^{n-2}(1+n \cdot 1)=n \cdot 2^{n-2}(1+n)=n(n+1) 2^{n-2}
\)
Step 5: Combine with \(\sum k C_k\)
\(
S=\sum k^2 C_k+\sum k C_k=n(n+1) 2^{n-2}+n 2^{n-1}=n(n+1) 2^{n-2}+n \cdot 2 \cdot 2^{n-2}=n(n+1+2) 2^{n-2}=n(n+3) 2^{n-2}
\)
Example 7: If \(n>2\), then prove that \(C_1(a-1)-C_2 \times (a-2)+\cdots+(-1)^{n-1} C_n(a-n)=a\), where \(C_r={ }^n C_r\).
Solution: The \(r\)-th term is:
\(
T_r=(-1)^{r-1}\binom{n}{r}(a-r)
\)
So the sum is:
\(
S=\sum_{r=1}^n(-1)^{r-1}\binom{n}{r}(a-r)
\)
Split the term \((a-r)\) into two parts
\(
\begin{aligned}
& S=\sum_{r=1}^n(-1)^{r-1}\binom{n}{r} a-\sum_{r=1}^n(-1)^{r-1}\binom{n}{r} r \\
& S=a \sum_{r=1}^n(-1)^{r-1}\binom{n}{r}-\sum_{r=1}^n(-1)^{r-1} r\binom{n}{r}
\end{aligned}
\)
Evaluate the first sum
We know:
\(
\sum_{r=0}^n(-1)^r\binom{n}{r}=(1-1)^n=0
\)
Thus:
\(
\sum_{r=1}^n(-1)^{r-1}\binom{n}{r}=-\sum_{r=1}^n(-1)^r\binom{n}{r}=1
\)
So the first part gives:
\(
a \sum_{r=1}^n(-1)^{r-1}\binom{n}{r}=a
\)
Evaluate the second sum
Use the identity \(r\binom{n}{r}=n\binom{n-1}{r-1}\) :
\(
\sum_{r=1}^n(-1)^{r-1} r\binom{n}{r}=\sum_{r=1}^n(-1)^{r-1} n\binom{n-1}{r-1}=n \sum_{r=1}^n(-1)^{r-1}\binom{n-1}{r-1}
\)
Change the index \(k=r-1\) :
\(
\sum_{r=1}^n(-1)^{r-1}\binom{n-1}{r-1}=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}=(1-1)^{n-1}=0
\)
So the second sum is 0.
Combine results
\(
S=a-0=a
\)
Example 8: Find the sum \(3{ }^n C_0-8^n C_1+13^n C_2-18 \times{ }^n C_3+\cdots \)
Solution: The pattern of coefficients is:
\(
3,-8,13,-18, \ldots
\)
Notice the coefficient of \(\binom{n}{r}\) is:
\(
(-1)^r(3+5 r)
\)
So the sum can be written as:
\(
S=\sum_{r=0}^n(-1)^r(3+5 r)\binom{n}{r}
\)
Split into two sums
\(
\begin{aligned}
S & =\sum_{r=0}^n(-1)^r \cdot 3\binom{n}{r}+\sum_{r=0}^n(-1)^r \cdot 5 r\binom{n}{r} \\
S & =3 \sum_{r=0}^n(-1)^r\binom{n}{r}+5 \sum_{r=0}^n(-1)^r r\binom{n}{r}
\end{aligned}
\)
Evaluate the first sum
\(
\sum_{r=0}^n(-1)^r\binom{n}{r}=(1-1)^n=0
\)
So the first sum contributes \(3 \cdot 0=0\).
Evaluate the second sum
Use the identity \(r\binom{n}{r}=n\binom{n-1}{r-1}\) :
\(
\sum_{r=0}^n(-1)^r r\binom{n}{r}=\sum_{r=1}^n(-1)^r n\binom{n-1}{r-1}=n \sum_{r=1}^n(-1)^r\binom{n-1}{r-1}
\)
Change index \(k=r-1\) :
\(
\sum_{r=1}^n(-1)^r\binom{n-1}{r-1}=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}=-\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}=-(1-1)^{n-1}=0
\)
So the second sum is \(5 \cdot 0=0\).
Combine results
\(
S=0+0=0
\)
Example 9: If \(x+y=1\), prove that \(\sum_{r=0}^n r^n C_r x^r y^{n-r} =n x\).
Solution: Factor out \(r\) using the identity \(r\binom{n}{r}=n\binom{n-1}{r-1}\)
\(
\begin{gathered}
\sum_{r=0}^n r\binom{n}{r} x^r y^{n-r}=\sum_{r=1}^n n\binom{n-1}{r-1} x^r y^{n-r} \\
\quad=\sum_{r=1}^n n\binom{n-1}{r-1} x \cdot x^{r-1} y^{(n-1)-(r-1)} \\
\quad=n x \sum_{r=1}^n\binom{n-1}{r-1} x^{r-1} y^{(n-1)-(r-1)}
\end{gathered}
\)
Recognize the binomial expansion
\(
\sum_{r=1}^n\binom{n-1}{r-1} x^{r-1} y^{(n-1)-(r-1)}=\sum_{k=0}^{n-1}\binom{n-1}{k} x^k y^{(n-1)-k}=(x+y)^{n-1}
\)
Substitute \(x+y=1\)
\(
n x(x+y)^{n-1}=n x \cdot 1^{n-1}=n x
\)
\(
\sum_{r=0}^n r\binom{n}{r} x^r y^{n-r}=n x
\)
Example 10: If \((1+x)^n=\sum_{r=0}^n C_r x^r\), show that
\(
C_0+\frac{C_1}{2}+\cdots+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}
\)
Solution:
\(
\begin{aligned}
&r^{\text {th }} \text { term of the given series is }\\
&T_r=\frac{{ }^n C_{r-1}}{r}=\frac{{ }^{n+1} C_r}{n+1}
\end{aligned}
\)
Required sum is
\(
\begin{aligned}
\sum_{r=1}^{n+1} T_r & =\sum_{r=1}^{n+1} \frac{{ }^{n+1} C_r}{n+1} \\
& =\frac{1}{n+1}\left({ }^{n+1} C_1+{ }^{n+1} C_2+\cdots+{ }^{n+1} C_{n+1}\right)
\end{aligned}
\)
\(
=\frac{2^{n+1}-1}{n+1}
\)
Example 11: Prove that \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\cdots=\frac{2^n-1}{n+1}\) where \(C_r={ }^n C_r\).
Solution:
\(
S=\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\cdots
\)
General term of the series is
\(
\frac{{ }^n C_{2 r-1}}{2 r}=\frac{{ }^{n+1} C_{2 r}}{n+1}
\)
where \(r=1,2,3, \ldots\)
\(
\begin{aligned}
\therefore \quad S & =\frac{1}{n+1}\left[{ }^{n+1} C_2+{ }^{n+1} C_4+{ }^{n+1} C_6+\cdots\right] \\
& =\frac{1}{n+1}\left[\left({ }^{n+1} C_0+{ }^{n+1} C_4+{ }^{n+1} C_6+\cdots\right)-{ }^{n+1} C_0\right] \\
& =\frac{1}{n+1}\left[2^n-1\right]
\end{aligned}
\)
Example 12: Find the sum \(2 C_0+\frac{2^2}{2} C_1+\frac{2^3}{3} C_2+\frac{2^4}{4} C_3+\cdots+\frac{2^{11}}{11} C_{10}\)
Solution: We have,
\(
\begin{aligned}
& 2 C_0+\frac{2^2}{2} C_1+\frac{2^3}{3} C_2+\cdots+\frac{2^{11}}{11} C_{10} \\
& =\sum_{r=0}^{10}{ }^{10} C_r \frac{2^{r+1}}{r+1} \\
& =\frac{1}{11} \sum_{r=0}^{10} \frac{11}{r+1}{ }^{10} C_r 2^{r+1} \\
& =\frac{1}{11} \sum_{r=0}^{10}{ }^{11} C_{r+1} 2^{r+1} \\
& =\frac{1}{11}\left({ }^{11} C_1 2^1+\cdots+{ }^{11} C_{11} 2^{11}\right) \\
& =\frac{1}{11}\left({ }^{11} C_0 2^0+{ }^{11} C_1 2^1+\cdots+{ }^{11} C_{11} 2^{11}-{ }^{11} C_0 2^0\right)
\end{aligned}
\)
\(
=\frac{1}{11}\left[(1+2)^{11}-1\right]=\frac{3^{11}-1}{11}
\)
Example 13: Prove that \(\sum_{r=0}^n{ }^n C_r \sin r x \cos (n-r) x =2^{n-1} \sin (n x)\).
Solution: Here sum is given by
\(
\begin{aligned}
& S=\sum_{r=0}^n{ }^n C_r \sin r x \cos (n-r) x \\
\Rightarrow & \left.S=\sum_{r=0}^n{ }^n C_{n-r} \sin (n-r) x \cos r x \text { (replacing } r \text { by } n-r\right) \\
\Rightarrow & 2 S=\sum_{r=0}^n{ }^n C_r \sin n x=\sin n x \times 2^n \\
\Rightarrow & S=2^{n-1} \sin n x
\end{aligned}
\)
Example 14: If for \(n \in N, \sum_{k=0}^{2 n}(-1)^k\left({ }^{2 n} C_k\right)^2=A\), then find the value of \(\sum_{k=0}^{2 n}(-1)^k(k-2 n)\left({ }^{2 n} C_k\right)^2\).
Solution: Let,
\(
S=\sum_{k=0}^{2 n}(-1)^k(k-2 n)\left({ }^{2 n} C_k\right)^2 \dots(1)
\)
\(
\Rightarrow \quad S=-\sum_{k=0}^{2 n}(-1)^{2 n-k} \cdot(2 n-k)\left({ }^{2 n} C_{2 n-k}\right)^2
\)
Writing the terms in \(S\) in the reverse order, we get
\(
S=-\sum_{k=0}^{2 n}(-1)^k\left({ }^{2 n} C_k\right)^2 \dots(2)
\)
Adding (1) and (2), we get
\(
\begin{aligned}
2 S & =-2 n \sum_{k=0}^{2 n}(-1)^k\left({ }^{2 n} C_k\right)^2=-2 n A \\
\Rightarrow \quad S & =-n A
\end{aligned}
\)
Series from Multiplication of Two Series
Case-I: \({ }^m C_r+{ }^m C_{r-1}{ }^n C_1+{ }^m C_{r-2}{ }^n C_2+\cdots+{ }^n C_r={ }^{m+n} C_r\), where \(r<m, r<n\) and \(m, n, r\) are + ve integers.
Proof: Suppose we have \(m+n[latex] objects, divided into:
[latex]m[latex] objects of type A
[latex]n[latex] objects of type B
We want to choose [latex]r[latex] objects from all [latex]m+n[latex].
Right-hand side
[latex]
{ }^{m+n} C_r
[latex]
counts the total number of ways to choose [latex]r[latex] objects from all [latex]m+n[latex].
For example, Let [latex]m=5, \quad n=4, \quad r=2\)
LHS:
\(
\begin{gathered}
{ }^5 C_2+{ }^5 C_1{ }^4 C_1+{ }^4 C_2 \\
=10+(5)(4)+6 \\
=10+20+6=36
\end{gathered}
\)
RHS:
\(
{ }^9 C_2=\frac{9 \cdot 8}{2}=36
\)
Case-II: \({ }^n C_0^2+{ }^n C_1^2+\cdots+{ }^n C_n^2={ }^{2 n} C_n\)
Proof: \({ }^n C_0^2+{ }^n C_1^2+{ }^n C_2^2+\cdots+{ }^n C_n^2\)
\(
\begin{aligned}
& =\left({ }^n C_0{ }^n C_n+{ }^n C_1{ }^n C_{n-1}+{ }^n C_2{ }^n C_{n-2}+\cdots+{ }^n C_n{ }^n C_0\right) \\
& =\text { Coefficient of } x^n \text { in }(1+x)^n(1+x)^n \\
& =\text { Coefficient of } x^n \text { in }(1+x)^{2 n} \\
& ={ }^{2 n} C_n
\end{aligned}
\)
For example, Case where \(\boldsymbol{n} \boldsymbol{=} \mathbf{3}\)
We calculate both sides of the equation:
Left-Hand Side (LHS):
The sum of squares of binomial coefficients for \(n=3\) is:
\(
\left({ }^3 C_0\right)^2+\left({ }^3 C_1\right)^2+\left({ }^3 C_2\right)^2+\left({ }^3 C_3\right)^2
\)
Recall that \({ }^3 C_0=1,{ }^3 C_1=3,{ }^3 C_2=3\), and \({ }^3 C_3=1\).
\(
=(1)^2+(3)^2+(3)^2+(1)^2=1+9+9+1=20
\)
Right-Hand Side (RHS):
The binomial coefficient \({ }^{2 n} C_n\) for \(n=3\) is \({ }^{2 \times 3} C_3={ }^6 C_3\).
\(
{ }^6 C_3=\frac{6!}{3!(6-3)!}=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20
\)
Result: LHS = RHS (20 = 20).
Case-III: \({ }^n C_0^2-{ }^n C_1^2+{ }^n C_2^2+\cdots+(-1)^n{ }^n C_n^2\)
Proof:
\(
\begin{aligned}
& { }^n C_0^2-{ }^n C_1^2+{ }^n C_2^2-\cdots+(-1)^n{ }^n C_n^2 \\
& ={ }^n C_0{ }^n C_n-{ }^n C_1{ }^n C_{n-1}+{ }^n C_2{ }^n C_{n-2}+\cdots+(-1)^n{ }^n C_n{ }^n C_0 \\
& =\text { Coefficient of } x^n \text { in }(1+x)^n(1-x)^n \\
& =\text { Coefficient of } x^n \text { in }\left(1-x^2\right)^n
\end{aligned}
\)
Now, general term in \(\left(1-x^2\right)^n\) is \(T_{r+1}={ }^n C_r\left(-x^2\right)^r\).
For \(x^{\prime \prime}, 2 r=n \Rightarrow r=n / 2 \Rightarrow n\) must be even.
If \(n\) is odd, \(r\) is not an integer or we can say, when \(n\) is odd, \(x^n\) term does not occur in \(\left(1-x^2\right)^n\) which is obvious. When \(r\) is even, \(r=n / 2\). Hence,
\(
\begin{aligned}
& T_{n / 2+1}={ }^n C_{n / 2}\left(-x^2\right)^{n / 2}=(-1)^n C_{n / 2} x^n \\
\Rightarrow & C_0^2-C_1^2+C_2^2+\cdots+(-1)^n C_n^2
\end{aligned}
\)
\(
=\left\{\begin{array}{cl}
0, & \text { if } n \text { is odd } \\
(-1)^n{ }^n C_{n / 2}, & \text { if } n \text { is even }
\end{array}\right.
\)
For example, \(n=3\) (odd)
\(
\begin{gathered}
\binom{3}{0}^2-\binom{3}{1}^2+\binom{3}{2}^2-\binom{3}{3}^2 \\
=1-9+9-1=0
\end{gathered}
\)
Formula gives: \(0\)
For example, \(n=4\) (even)
\(
\begin{gathered}
\binom{4}{0}^2-\binom{4}{1}^2+\binom{4}{2}^2-\binom{4}{3}^2+\binom{4}{4}^2 \\
=1-16+36-16+1=6
\end{gathered}
\)
Formula gives:
\(
(-1)^2\binom{4}{2}=6
\)
Example 15:Â Find the sum \(\sum_{i=0}^r{ }^{n_1} C_{(r-i)}{ }^{n_2} C_i\).
Solution: Let
\(
S=\sum_{i=0}^r\binom{n_1}{r-i}\binom{n_2}{i} .
\)
From the binomial theorem,
\(
(1+x)^{n_1}=\sum_{k=0}^{n_1}\binom{n_1}{k} x^k, \quad(1+x)^{n_2}=\sum_{j=0}^{n_2}\binom{n_2}{j} x^j .
\)
The coefficient of \(x^r\) in the product
\(
(1+x)^{n_1}(1+x)^{n_2}
\)
is
\(
\sum_{i=0}^r\binom{n_1}{r-i}\binom{n_2}{i} .
\)
But
\(
(1+x)^{n_1}(1+x)^{n_2}=(1+x)^{n_1+n_2}
\)
whose coefficient of \(x^r\) is
\(
\binom{n_1+n_2}{r}
\)
\(
\sum_{i=0}^r\binom{n_1}{r-i}\binom{n_2}{i}=\binom{n_1+n_2}{r}
\)
For example, Given: \(n_1=3, n_2=2, r=2\)
\(
\begin{gathered}
\sum_{i=0}^2\binom{3}{2-i}\binom{2}{i}=\binom{3}{2}\binom{2}{0}+\binom{3}{1}\binom{2}{1}+\binom{3}{0}\binom{2}{2} \\
=3 \cdot 1+3 \cdot 2+1 \cdot 1=3+6+1=10
\end{gathered}
\)
RHS:
\(
\binom{3+2}{2}=\binom{5}{2}=10
\)
Example 16:Â Prove that \(\sum_{i=0}^{2 n} r\left({ }^{2 n} C_r\right)^2={ }^{4 n-1} C_{2 n-1}\).
Solution:
\(
\begin{aligned}
S & =\sum_{r=0}^{2 n} r\left({ }^{2 n} C_r\right)^2 \\
& =\sum_{r=0}^{2 n}\left(r^{2 n} C_r\right)\left({ }^{2 n} C_r\right) \\
& =\sum_{r=0}^{2 n}(2 n)^{2 n-1} C_{r-1}{ }^{2 n} C_r \\
& =2 n \sum_{r=0}^{2 n}{ }^{2 n-1} C_{r-1}{ }^{2 n} C_{2 n-r}
\end{aligned}
\)
Here, the sum of suffixes is \(r-1+(2 n-r)=2 n-1\) which is constant.
\(
\begin{aligned}
\therefore \quad S= & \text { Coefficient of } x^{2 n-1} \text { in the expansion of } \\
& (1+x)^{2 n-1}(1+x)^{2 n} \\
= & \text { Coefficient of } x^{2 n-1} \text { in the expansion of }(1+x)^{4 n-1} \\
= & { }^{4 n-1} C_{2 n-1}
\end{aligned}
\)
Example 17: \(A\) is a set containing \(n\) elements. A subset \(P\) of \(A\) is chosen at random. The set \(A\) is reconstructed by replacing the elements of \(P\). A subset \(Q\) is again chosen at random. The number of ways of selecting \(P\) and \(Q\), is
(a) \(2^n\)
(b) \(4^n\)
(c) \(2 n\)
(d) \(3^n\)
solution: (b) The set \(A\) has \(n\) elements. So, it has \(2^n\) subsets. Therefore, set \(P\) can be chosen in \({ }^{2^n} C_1\) ways. Similarly, set \(Q\) can also be chosen in \({ }^{2^n} C_1\) ways. ∴ Sets \(P\) and \(Q\) can be chosen \(\operatorname{in}^{2^n} C_1 \times 2^n C_1=2^n \times 2^n=4^n\) ways.
Binomial Theorem for any Index
Let \(n\) be a rational number and \(x\) be a real number such that \(|x|<1\), then
\(
\begin{aligned}
(1+x)^n=1+n x & +\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3 \\
& +\ldots+\frac{n(n-1)(n-2) \ldots(n-r+1)}{r!} x^r+\ldots \infty
\end{aligned}
\)
Remark 1: The condition \(|x|<1\) is un-necessary, if \(n\) is a whole number while the same condition is essential if \(n\) is a rational number other than a whole number.
Remark 2: Note that there are infinite number of terms in the expansion of \((1+x)^n\), when \(n\) is a negative integer or a fraction.
Remark 3: In the above expansion the first term is unity. If the first term is not unity and the index of the binomial is either a negative integer or a fraction, then we expand as follows:
\(
(x+a)^n=\left\{a\left(1+\frac{x}{a}\right)\right\}^n=a^n\left(1+\frac{x}{a}\right)^n
\)
\(
\Rightarrow \quad(x+a)^n=a^n\left\{1+n \frac{x}{a}+\frac{n(n-1)}{2!}\left(\frac{x}{a}\right)^2+\ldots\right\}
\)
\(
\Rightarrow \quad(x+a)^n=a^n+n a^{n-1} x+\frac{n(n-1)}{2!} a^{n-2} x^2+\ldots
\)
This expansion is valid when \(\left|\frac{x}{a}\right|<1\) or equivalently \(|x|<|a|\).
Remark 4: If \(n\) is a positive integer the above expansion contains \((n+1)\) terms and coincides with
\(
(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots+{ }^n C_n x^n .
\)
General term
he general term in the expansion of \((1+x)^n\) is given by
\(
T_{r+1}=\frac{n(n-1)(n-2) \ldots(n-(r-1))}{r!} x^r
\)
Important Expansions
Example 18: Find the condition for which the formula
\(
(a+b)^m=a^m+m a^{m-1} b+\frac{m(m-1)}{1 \times 2} a^{m-2} b^2+\cdots \text { holds. }
\)
Solution: The expression can be written as \(a^m\left\{\left(1+\frac{b}{a}\right)^m\right\}\). Hence, it is valid only when
\(
\begin{aligned}
& |b / a|<1 \\
\Rightarrow & |b|<|a|
\end{aligned}
\)
Note: In mathematics, when an expansion converges, it means that the sum of its infinite terms approaches and ultimately settles at a specific, finite value, known as its limit.
Example 19: Find the sum
\(
1-\frac{1}{8}+\frac{1}{8} \times \frac{3}{16}-\frac{1 \times 3 \times 5}{8 \times 16 \times 24}+\ldots
\)
Solution: Comparing the given series to
\(
1+n x+\frac{n(n-1)}{2!} x^2+\cdots=(1+x)^n
\)
we get
\(
n x=-\frac{1}{8} \text { and } \frac{n(n-1)}{2!} x^2=\frac{3}{128}
\)
\(
\Rightarrow \quad x=\frac{1}{4}, n=-\frac{1}{2}
\)
Hence,
\(
1-\frac{1}{8}+\frac{1}{8} \frac{3}{16}-\ldots=\left(1+\frac{1}{4}\right)^{-1 / 2} \cdot=\frac{2}{\sqrt{5}}
\)
Example 20: Find the coefficient of \(x^n\) in the expansion of \(\left(1-9 x+20 x^2\right)^{-1}\).
Solution: We have,
\(
\begin{aligned}
& \left(1-9 x+20 x^2\right)^{-1}=[(1-5 x)(1-4 x)]^{-1} \\
& =\frac{1}{(1-5 x)(1-4 x)}=\frac{5}{1-5 x}-\frac{4}{1-4 x} \\
& =5(1-5 x)^{-1}-4(1-4 x)^{-1}
\end{aligned}
\)
\(
\begin{array}{r}
=5\left[1+5 x+(5 x)^2+\cdots+(5 x)^n+\cdots\right]-4\left[1+4 x+(4 x)^2\right. \\
\left.+\cdots+(4 x)^n+\cdots\right]
\end{array}
\)
Therefore the coefficient of \(x^n\) is \(5^{n+1}-4^{n+1}\).
Example 21: In the binomial expansion of \((a-b)^n, n \geq 5\), the sum of the \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero. Then, \(a / b\) equals [IIT (S) 2001, AIEEE 2007]
(a) \(\frac{n-5}{6}\)
(b) \(\frac{n-4}{5}\)
(c) \(\frac{5}{n-4}\)
(d) \(\frac{6}{n-5}\)
Solution: (b) We have,
\(
\begin{aligned}
& T_5+T_6=0 \\
\Rightarrow & { }^n C_4 a^{n-4} b^4-{ }^n C_5 a^{n-5} b^5=0 \\
\Rightarrow & { }^n C_4 a={ }^n C_5 b \Rightarrow \frac{a}{b}=\frac{{ }^n C_5}{{ }^n C_4} \Rightarrow \frac{a}{b}=\frac{n-4}{5}
\end{aligned}
\)
Example 22: The coefficient of the middle term in the binomial expansion, in powers of \(x\), of \((1+\alpha x)^4\) and of \((1-\alpha x)^6\) is same, if \(\alpha\) equals [AIEEE 2004]
(a) \(\frac{3}{5}\)
(b) \(\frac{10}{3}\)
(c) \(-\frac{3}{10}\)
(d) \(-\frac{5}{3}\)
Solution: (c) We have,
In the expansion of \((1+\alpha x)^4\),
\(
\text { Middle term }={ }^4 C_2(\alpha x)^2=6 \alpha^2 x^2
\)
In the expansion of \((1-\alpha x)^6\),
\(
\text { Middle term }={ }^6 C_3(-\alpha x)^3=-20 \alpha^3 x^3
\)
\(\therefore \quad\) Coefficient of the middle term in \((1+\alpha x)^4\)
\(=\) Coefficient of the middle term in \((1-\alpha x)^6\)
\(
\Rightarrow \quad 6 \alpha^2=-20 \alpha^3 \Rightarrow \alpha=0, \alpha=-\frac{3}{10}
\)
Example 23: Coefficient of \(t^{24}\) in \(\left(1-t^2\right)^{12}\left(t+t^{12}\right)\left(1-t^{24}\right)\), is [IIT (S) 2002]
(a) \({ }^{12} C_6+3\)
(b) \({ }^{12} C_6+1\)
(c) \({ }^{12} C_6\)
(d) \({ }^{12} C_6+2\)
Solution: (c) We have,
\(
\begin{aligned}
& (1-t)^{12}\left(t+t^{12}\right)\left(1-t^{24}\right) \\
& =\left\{\sum_{r=0}^{12}{ }^{12} C_r\left(-t^2\right)^r\right\}\left\{t-t^{25}+t^{12}-t^{36}\right\} \\
& =\sum_{r=0}^{12}{ }^{12} C_r\left(-t^2\right)^r\left\{t+t^{12}-t^{25}-t^{36}\right\}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \quad \text { Coefficient of } t^{24} & =\text { Coefficient of } t^{12} \text { in } \sum_{r=0}^{12} C_r\left(-t^2\right)^r \\
& ={ }^{12} C_6
\end{aligned}
\)
Example 24: If \(n\) is an odd natural number, then \(\sum_{r=0}^n \frac{(-1)^r}{{ }^n C_r}\) equals [IIT 1998]
(a) 0
(b) \(\frac{1}{n}\)
(c) \(\frac{n}{2^n}\)
(d) none of these
Solution: (a) We have,
\(
\begin{aligned}
& \left.\left.\sum_{r=0}^n \frac{(-1)^r}{{ }^n C_r}=\sum_{r=0}^{\frac{n+1}{2}} \right\rvert\, \frac{(-1)^r}{{ }^n C_r}+\frac{(-1)^{n-r}}{{ }^n C_{n-r}}\right\} \\
\Rightarrow & \sum_{r=0}^n \frac{(-1)^r}{{ }^n C_r}=\sum_{r=0}^{\frac{n+1}{2}}(-1)^r\left\{\frac{1}{{ }^n C_r}+\frac{(-1)^n}{{ }^n C_{n-r}}\right\} \\
\Rightarrow & \sum_{r=0}^n \frac{(-1)^r}{{ }^n C_r}=\sum_{r=0}^{\frac{n+1}{2}}(-1)^2\left\{\frac{1}{{ }^n C_r}-\frac{1}{{ }^n C_r}\right\}=0\left[\because \begin{array}{l}
n \text { is odd and } \\
{ }^n C_r={ }^n C_{n-r}
\end{array}\right]
\end{aligned}
\)
Example 25: If \(s_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}\) and \(t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}\), then \(\frac{t_n}{s_n}\) is equal to [AIEEE 2004]
(a) \(\frac{2 n-1}{2}\)
(b) \(\frac{n}{2}-1\)
(c) n-1
(d) \(\frac{n}{2}\)
Solution: (d) The key to solving this problem is the symmetry property of binomial coefficients, which states that \({ }^n C_r={ }^n C_{n-r}\).
Express \(\boldsymbol{t}_{\boldsymbol{n}}\) in two ways and combine
The expression for \(t_n\) is given by \(t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}\). We can rewrite this sum by replacing \(r\) with \(n-r\) and using the symmetry property:
\(
t_n=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}=\sum_{r=0}^n \frac{n-r}{{ }^n C_r}
\)
Now, we add the original expression for \(t_n\) and this new form:
\(
\begin{gathered}
t_n+t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}+\sum_{r=0}^n \frac{n-r}{{ }^n C_r} \\
2 t_n=\sum_{r=0}^n \frac{r+(n-r)}{{ }^n C_r}=\sum_{r=0}^n \frac{n}{{ }^n C_r} \\
2 t_n=n \sum_{r=0}^n \frac{1}{{ }^n C_r} \dots(1)
\end{gathered}
\)
Relate to \(s_n\) and find the ratio
Recognizing that \(s_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}\), the equation (1) becomes:
\(
2 t_n=n s_n
\)
Therefore, the ratio \(\frac{t_n}{s_n}\) is:
\(
\frac{t_n}{s_n}=\frac{n}{2}
\)
Example 26: The remainder left out when \(8^{2 n}-(62)^{2 n+1}\) is divided by 9, is [AIEEE 2009]
(a) 0
(b) 2
(c) 7
(d) 8
Solution: (b) We have,
\(
\begin{aligned}
& 8^{2 n}-(62)^{2 n+1} \\
&=(63+1)^n-(63-1)^{2 n+1} \\
&=(1+63)^n+(1-63)^{2 n+1} \\
&=\left\{1+{ }^n C_1 \times 63+{ }^n C_2 \times(63)^2+\ldots+{ }^n C_n \times(63)^n\right\} \\
&+ 1-{ }^{2 n+1} C_1 \times 63+{ }^{2 n+1} C_2 \times(63)^2+\ldots+(-1)^{2 n+1}(63)^{2 n+1} \\
&= 2+63\left\{{ }^n C_1+{ }^n C_2 \times 63+\ldots+{ }^n C_n \times(63)^{n-1}\right\} \\
&+\left\{-{ }^{2 n+1} C_1+{ }^{2 n+1} C_2 \times 63+\ldots+(-1)^{2 n+1}(63)^{2 n}\right\} \times 63 \\
&= 2+\text { A multiple of } 63 \\
& \text { Required remainder }=2
\end{aligned}
\)
Alternate: Step 1: Use modulo 9 equivalence
Notice that:
\(
8 \equiv-1(\bmod 9), \quad 62 \equiv-1(\bmod 9) \quad(\text { since } 62=9 \cdot 6+8)
\)
So:
\(
8^{2 n} \equiv(-1)^{2 n} \equiv 1 \quad(\bmod 9), \quad 62^{2 n+1} \equiv(-1)^{2 n+1} \equiv-1 \quad(\bmod 9)
\)
Step 2: Subtract
\(
8^{2 n}-62^{2 n+1} \equiv 1-(-1) \equiv 2 \quad(\bmod 9)
\)
Example 27: For natural numbers \(m, n\) if \((1-y)^m(1+y)^n=1+a_1 y+a_2 y^2+\ldots\) and \(a_1=a_2=10\), then ( \(m, n\) ), is [AIEEE 2006]
(a) \((20,45)\)
(b) \((35,20)\)
(c) \((45,35)\)
(d) \((35,45)\)
Solution: (d) We have
\(
\begin{array}{ll}
& (1-y)^m(1+y)^n \\
& =\left({ }^m C_0-{ }^m C_1 y+{ }^m C_2 y^2-\ldots\right) \times\left({ }^n C_0+{ }^n C_1 y+{ }^n C_2 y^2+\ldots\right) \\
& =1+(n-m) y+\left\{\frac{m(m-1)+n(n-1)}{2}-m n\right\} y^2+\ldots \\
\therefore \quad & a_1=n-m \text { and } a_2=\frac{m(m-1)+n(n-1)}{2}-m n \\
\Rightarrow \quad & n-m=10 \text { and } \frac{m^2+n^2-(m+n)-2 m n}{2}=10 \\
\Rightarrow \quad & \quad n-m=10 \text { and }(m-n)^2-(m+n)=20 \\
\Rightarrow \quad & m-n=-10 \text { and } m+n=80 \\
\Rightarrow \quad & m=35 \text { and } n=45
\end{array}
\)
Example 28: If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 a^3+\ldots\), then \(a_n\), is [AIEEE 2006]
(a) \(\frac{b^n-a^n}{b-a}\)
(b) \(\frac{a^n-b^n}{b-a}\)
(c) \(\frac{a^{n+1}-b^{n+1}}{b-a}\)
(d) \(\frac{b^{n+1}-a^{n+1}}{b-a}\)
Solution: (d) We have
\(
\begin{aligned}
& \frac{1}{(1-a x)(1-b x)}=\frac{1}{a-b}\left\{\frac{a}{1-a x}-\frac{b}{1-b x}\right\} \\
\Rightarrow & \frac{1}{(1-a x)(1-b x)}=\frac{1}{a-b}\left\{a(1-a x)^{-1}-b(1-b x)^{-1}\right\} \\
\Rightarrow & \frac{1}{(1-a x)(1-b x)}=\frac{1}{a-b}\left\{a \sum_{r=0}^{\infty}(a r)^r-b \sum_{r=0}^{\infty}(b x)^r\right\} \\
\Rightarrow & \frac{1}{(1-a x)(1-b x)}=\frac{1}{a-b}\left\{a\left(\sum_{r=0}^{\infty} a^r x^r\right)-b\left(\sum_{r=0}^{\infty} b^r x^r\right)\right\} \\
\therefore & a_n=\text { Coefficient of } x^n \text { in } \frac{1}{(1-a x)(1-b x)} \\
\Rightarrow & a_n=\frac{1}{a-b}\left\{a \times a^n-b \times b^n\right\}=\frac{a^{n+1}-b^{n+1}}{a-b}
\end{aligned}
\)
Example 29: \(\binom{30}{0}\binom{30}{10}-\binom{30}{1}\binom{30}{11}+\ldots+\binom{30}{20}\binom{30}{30}=\) [IIT (S) 2005]
(a) \({ }^{30} C_{11}\)
(b) \({ }^{60} \mathrm{C}_{10}\)
(c) \({ }^{30} \mathrm{C}_{10}\)
(d) \({ }^{65} \mathrm{C}_{55}\)
Solution: (c) We have
\(
\begin{aligned}
& { }^{30} C_0 \times{ }^{30} C_{10}-{ }^{30} C_1 \times{ }^{30} C_{11}+\ldots+{ }^{30} C_{10} \times{ }^{30} C_0 \\
& =\text { Coefficient of } x^{20} \text { in }\left\{(1+x)^{30}(1-x)^{30}\right\} \\
& =\text { Coefficient of } x^{20} \text { in }\left(1-x^2\right)^{30} \\
& ={ }^{30} C_{10}(-1)^{10}={ }^{30} C_{10}
\end{aligned}
\)
Example 30: The sum of the series [AIEEE 2007]
\(
{ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots+{ }^{20} C_{10} \text {, is }
\)
(a) 0
(b) \({ }^{20} \mathrm{C}_{10}\)
(c) \(-{ }^{20} C_{10}\)
(d) \(\frac{1}{2}{ }^{20} \mathrm{C}_{10}\)
Solution: (d) We have
\(
\begin{array}{r}
{ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots+{ }^{20} C_{10}-{ }^{20} C_{11}+\ldots \\
-{ }^{20} C_{19}+{ }^{20} C_{20}=0 \\
\Rightarrow \quad 2\left({ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots-{ }^{20} C_9+{ }^{20} C_{10}\right)-{ }^{20} C_{10}=0 \\
{\left[\because{ }^n C_r={ }^n C_{n-r}\right]} \\
\Rightarrow \quad{ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots-{ }^{20} C_9+{ }^{20} C_{10}=\frac{1}{2}{ }^{20} C_{10}
\end{array}
\)
Example 31: For \(r=0,1,2, \ldots, 10\), let \(A_r, B_r\) and \(C_r\) denote, respectively, the coefficient of \(x^r\) in the expansions of \((1+x)^{10},(1+x)^{20}\) and \((1+x)^{30}\). Then, \(\sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right)\) is equal to [IIT 2010]
(a) \(B_{10}-C_{10}\)
(b) \(A_{10}\left(B_{10}^2-C_{10} A_{10}\right)\)
(c) 0
(d) \(C_{10}-B_{10}\)
Solution: (d) We have,
\(
\begin{aligned}
& A_r={ }^{10} C_r, B_r={ }^{20} C_r \text { and } C_r={ }^{30} C_r \\
\therefore \quad & \sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right) \\
& =B_{10} \sum_{r=1}^{10} A_r B_r-C_{10} \sum_{r=1}^{10}\left(A_r\right)^2 \\
& =B_{10}\left(\sum_{r=1}^{10}{ }^{10} C_r{ }^{20} C_r\right)-C_{10} \sum_{r=1}^{10}\left({ }^{10} C_r\right)^2
\end{aligned}
\)
\(
\begin{aligned}
& =B_{10}\left\{\sum_{r=1}^{10}{ }^{10} C_r{ }^{20} C_{20-r}\right)-C_{10}\left\{\sum_{r=0}^{10}\left({ }^{10} C_r\right)^2-1\right\} \\
& =B_{10}\left\{\sum_{r=0}^{10}{ }^{10} C_r{ }^{20} C_{20-r}-1\right\}-C_{10}\left\{\sum_{r=0}^{10}\left({ }^{10} C_r\right)^2-1\right\} \\
& =B_{10} \times\left\{\text { Coefficient of } x^{20} \text { in }(1+x)^{10}(x+1)^{20}-1\right\} \\
& =B_{10} \times\left\{{ }^{30} C_{20}-1\right\}-C_{10}\left\{{ }^{20} C_{10}-1\right\} \quad-C_{10}\left\{{ }^{20} C_{10}-1\right\} \\
& =B_{10}\left\{{ }^{30} C_{10}-1\right\}-C_{10}\left\{{ }^{20} C_{10}-1\right\} \\
& =B_{10}\left(C_{10}-1\right)-C_{10}\left(B_{10}-1\right)=C_{10}-B_{10}
\end{aligned}
\)
Example 32: The coefficeint of \(x^7\) in the expansion of \(\left(1-x-x^2+x^3\right)^6\) is: [AIEEE 2011]
(a) 144
(b) -132
(c) -144
(d) 132
Solution: (c) We have,
\(
\left(1-x-x^2+x^3\right)^6=\left\{(1-x)\left(1-x^2\right)\right\}^7=(1-x)^7\left(1-x^2\right)^7
\)
\(
\begin{aligned}
& =\left(6 C_0-{ }^6 C_1 x+{ }^6 C_2 x^2-{ }^6 C_3 x^3+\ldots+{ }^6 C_6 x^6\right) \\
& \left({ }^6 C_0-{ }^6 C_1 x^2+{ }^6 C_2 x^4-{ }^6 C_3 x^6+\ldots\right) \\
& \therefore \quad \text { Coefficient of } x^7 \text { in }\left(1-x-x^2+x^3\right)^6 \\
& =-{ }^6 C_1 \times-{ }^6 C_3-{ }^6 C_3 \times{ }^6 C_2-{ }^6 C_5 \times-{ }^6 C_1 \\
& =120-300+36=-144
\end{aligned}
\)
Example 33: Let \(r>1\) and \(n>2\) be integers. Suppose \(L\) and \(M\) are the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms respectively in the binomial expansion of \(2 n(1+x)^{2 n-1}\). If \((r+2) L=3 r(M)\), then \(n\) is [AIEEE 2012]
(a) \(2 r-1\)
(b) \(2 r\)
(c) \(2 r+1\)
(d) \(2 r+2\)
Solution: (c) Clearly
\(
\begin{aligned}
& L=2 n \times{ }^{2 n-1} C_{3 r-1} \text { and } M=2 n^{2 n-1} C_{r+1} \\
\therefore \quad & (r+2) L=3 r(M) \\
\Rightarrow \quad & (r+2) 2 n \times{ }^{2 n-1} C_{3 r-1}=3 r \times 2 n^{2 n-1} C_{r+1}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{(2 n-1)!(r+2)}{(3 r-1)!(2 n-3 r)!}=\frac{3 r \times(2 n-1)!}{(2 n-r-2)!(r+1)!} \\
& \Rightarrow \quad \frac{1}{3 r!(2 n-3 r)!}=\frac{1}{(r+2)!(2 n-r-2)!} \\
& \Rightarrow \quad \frac{(2 n)!}{3 r!(2 n-3 r)!}=\frac{(2 n)!}{(r+2)!(2 n-r-2)!} \\
& \Rightarrow \quad{ }^{2 n} C_{3 r}={ }^{2 n} C_{r+2} \\
& \Rightarrow \quad 3 r+r+2=2 n \Rightarrow n=2 r+1
\end{aligned}
\)
Example 34: The coefficients of three consecutive terms of \((1+x)^{n+5}\) are in the ratio \(5: 10: 14\). Then, \(n=\) [JEE (Adv) 2013]
(a) 5
(b) 7
(c) 6
(d) 8
Solution: (c)
Let \(r^{\text {th }},(r+1)^{\text {th }}\) and \((r+2)^{\text {th }}\) be three consecutive terms in the expansion of \((1+x)^{n+5}\). It is given that
\(
\begin{aligned}
& { }^{n+5} C_{r-1}:{ }^{n+5} C_r:{ }^{n+5} C_{r+1}=5: 10: 14 \\
\Rightarrow & \frac{{ }^{n+5} C_r}{{ }^{n+5} C_{r-1}}=\frac{10}{5} \text { and } \frac{{ }^{n+5} C_{r+1}}{{ }^{n+5} C_r}=\frac{14}{10} \\
\Rightarrow & \frac{n+5-r+1}{r}=2 \text { and } \frac{n+5-r}{r+1}=\frac{7}{5} \\
\Rightarrow & n-3 r+6=0 \text { and } 5 n-12 r+18=0 \\
\Rightarrow & n=6, r=4
\end{aligned}
\)
Sum of the Series when \(i\) and \(j\) are Dependent
Consider, sum of the series \(\sum_{0 \leq i<j \leq n} f(i) f(j)\)
In the given summation, \(i\) and \(j\) are not independent.
In the sum of series \(\sum_{i=1}^n \sum_{j=1}^n f(i) f(j)=\sum_{i=1}^n\left(f(i)\left(\sum_{j=1}^n f(j)\right)\right)\), \(i\) and \(j\) are independent.
For example: \(n=2\)
Binomial coefficients: \({ }^2 C_0=1,{ }^2 C_1=2,{ }^2 C_2=1\)
All pairs \((i, j)\) with \(i<j\) :
\(
(0,1),(0,2),(1,2)
\)
\(
\begin{aligned}
S & =(1+2)+(1+1)+(2+1) \\
& =3+2+3=8
\end{aligned}
\)
Check formula: \(n \cdot 2^n=2 \cdot 2^2=2 \cdot 4=8\)
Example 35: Find the value of \({\sum \sum}_{0 \leq i<j \leq n}\left({ }^n C_i+{ }^n C_j\right) .\)
Solution:
\(
{\sum \sum}_{0 \leq i<j \leq n} \left({ }^n C_i+{ }^n C_j\right)
\)
\(
=\frac{\left(\sum_{i=0}^n \sum_{j=0}^n\left({ }^n C_i+{ }^n C_j\right)\right)-\sum_{i=0}^n 2{ }^n C_i}{2}
\)
\(
=\frac{\left(\sum_{i=0}^n\left(\sum_{j=0}^n{ }^n C_i+\sum_{j=0}^n{ }^n C_j\right)\right)-2 \times 2^n}{2}
\)
\(
\begin{aligned}
& =\frac{\left(\sum_{i=0}^n\left({ }^n C_i \sum_{j=0}^n 1+2^n\right)\right)-2^{n+1}}{2} \\
& =\frac{\left(\sum_{i=0}^n\left({ }^n C_i(n+1)+2^n\right)\right)-2^{n+1}}{2}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{(n+1) \sum_{i=1}^n{ }^n C_i+2^n \sum_{i=0}^n 1-2^{n+1}}{2} \\
& =\frac{(n+1) 2^n+2^n(n+1)-2^{n+1}}{2} \\
& =(n+1) 2^n-2^n=n 2^n
\end{aligned}
\)
Example 36: The larger of \(99^{50}+100^{50}\) and \(101^{50}\) is ____ [IIT-JEE, 1982]
Solution: We have,
\(
101^{50}=(100+1)^{50}=100^{50}+50 \times 100^{49}+\frac{50 \times 49}{2 \times 1} 100^{48}+\cdots \dots(1)
\)
\(
\begin{aligned}
& 99^{50}=(100-1)^{50}=100^{50}-50 \times 100^{49}+ \\
& \frac{50 \times 49}{2 \times 1} 100^{48}-\cdots \dots(2)
\end{aligned}
\)
Subtracting (2) from (1), we get
\(
101^{50}-99^{50}=100^{50}+2 \frac{50 \times 49 \times 48}{1 \times 2 \times 3} 100^{47}+\cdots>100^{50}
\)
Hence, \(101^{50}>100^{50}+99^{50}\).
Example 37: The sum of the coefficients of the polynomial \(\left(1+x-3 x^2\right)^{2163}\) is ____. [IIT-JEE, 1982]
Solution: The sum of the coefficients of a polynomial \(P(x)\) is:
\(
P(1)
\)
because when you substitute \(x=1\), each term contributes its coefficient to the sum.
Apply to the given polynomial
Given:
\(
P(x)=\left(1+x-3 x^2\right)^{2163}
\)
Sum of coefficients:
\(
P(1)=\left(1+1-3 \cdot 1^2\right)^{2163}=(2-3)^{2163}=(-1)^{2163}
\)
Simplify
\(
(-1)^{2163}=-1 \quad(\text { since } 2163 \text { is odd })
\)
Example 38: If \((1+a x)^n=1+8 x+24 x^2+\cdots\), then \(a=\) ____ and \(n=\) _____. [IIT-JEE, 1983]
Solution: We are given:
\(
(1+a x)^n=1+8 x+24 x^2+\cdots
\)
and we need to find \(a\) and \(n\).
Step 1: Recall the binomial expansion
\(
(1+a x)^n=1+n(a x)+\frac{n(n-1)}{2!}(a x)^2+\cdots
\)
Comparing with the given series:
\(
1+8 x+24 x^2+\cdots
\)
we get the coefficients term by term.
Step 2: Compare the coefficient of \(x\)
\(
n \cdot a=8 \quad \Rightarrow \quad a=\frac{8}{n}
\)
Step 3: Compare the coefficient of \(x^2\)
\(
\frac{n(n-1)}{2} a^2=24
\)
Substitute \(a=\frac{8}{n}\) :
\(
\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24
\)
\(
\begin{aligned}
&8 n=32 \quad \Rightarrow \quad n=4\\
&a=\frac{8}{n}=\frac{8}{4}=2
\end{aligned}
\)
Example 39: The sum of the rational terms in the expansion of \(\left(\sqrt{2}+3^{1 / 5}\right)^{10}\) is ____. [IIT-JEE, 1997]
Solution: The general term in the binomial expansion is:
\(
T_{r+1}=\binom{10}{r}(\sqrt{2})^{10-r}\left(3^{1 / 5}\right)^r=\binom{10}{r}(\sqrt{2})^{10-r} 3^{r / 5}
\)
where \(r=0,1,2, \ldots, 10\).
Step 2: Condition for rationality
For a term to be rational, the powers of \(\sqrt{2}\) and \(3^{1 / 5}\) must be integers.
\((\sqrt{2})^{10-r}=2^{(10-r) / 2} \rightarrow\) exponent must be integer: \((10-r) / 2\) is integer.
So \(10-r\) must be even, i.e., \(r\) is even.
\(3^r / 5 \rightarrow\) exponent must be integer: \(r / 5\) is integer \(\rightarrow r\) divisible by 5.
Step 3: Find valid \(r\)
\(r=0,2,4,6,8,10\) (even numbers)
\(r=0,5,10\) (multiples of 5 )
Intersection: \(r=0,10\) only.
Step 4: Compute the terms
\(r=0: T_1=\binom{10}{0}(\sqrt{2})^{10}\left(3^{1 / 5}\right)^0=(\sqrt{2})^{10}=2^5=32\)
\(r=10: T_{11}=\binom{10}{10}(\sqrt{2})^0\left(3^{1 / 5}\right)^{10}=\left(3^{1 / 5}\right)^{10}=3^2=9\)
Step 5: Sum of rational terms
\(
32+9=41
\)
Example 40: If \({ }^{n-1} C_r=\left(k^2-3\right)^n C_{r+1}\), then \(k \in\) [IIT-JEE, 2004]
a. \((-\infty,-2]\)
b. \([2, \infty)\)
c. \([-\sqrt{3}, \sqrt{3}]\)
d. \((\sqrt{3}, 2]\)
Solution:
\(
\begin{aligned}
& \text { 10. d. }{ }^{n-1} C_r={ }^n C_{r+1}\left(k^2-3\right) \\
& \Rightarrow k^2-3=\frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=\frac{r+1}{n}
\end{aligned}
\)
Now,
\(
\begin{aligned}
& 0 \leq r \leq n-1 \\
\Rightarrow & 1 \leq r+1 \leq n \\
\Rightarrow & \frac{1}{n} \leq \frac{r+1}{n} \leq 1 \\
\Rightarrow & \frac{1}{n} \leq k^2-3 \leq 1 \\
\Rightarrow & 3+\frac{1}{n} \leq k^2 \leq 4 \Rightarrow \sqrt{3+\frac{1}{n}} \leq k \leq 2
\end{aligned}
\)
When \(n \rightarrow \infty\), we have
\(
\begin{aligned}
& \sqrt{3}<k \leq 2 \\
\Rightarrow \quad & k \in(\sqrt{3}, 2]
\end{aligned}
\)
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