Binomial Theorem
The Binomial Theorem is a formula to expand powers of binomials (expressions with two terms like \(a+b\) ) into a sum of terms, revealing coefficients from Pascal’s triangle and showing how exponents decrease for the first term and increase for the second. It’s written as \((a+b)^n=\sum_{k=0}^n\binom{n}{k} a^{n-k} b^k\), providing a shortcut to expand \((a+b)^n\) without tedious multiplication, useful in probability, calculus, and engineering.
Note: The word “Binomial” comes from BI meaning two, and NOMIAL meaning terms. Thus, a binomial expression has two terms connected by operators ‘ + ‘ or ‘ – ‘.
The Formula
\((a+b)^n=\binom{n}{0} a^n+\binom{n}{1} a^{n-1} b+\binom{n}{2} a^{n-2} b^2+\ldots+\binom{n}{n} b^n\)
\((a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\cdots+{ }^n C_n b^n, \quad \text { Where } n \in N\)
Or using sigma notation: \((a+b)^n=\sum_{k=0}^n\binom{n}{k} a^{n-k} b^k\)
Note: \(\left(\begin{array}{c}n \\ k\end{array}\right)={ }^n C_k=\frac{(n)(n-1)(n-2) \cdots(n-(k-1))}{k !}=\frac{n !}{k !(n-k) !}\).
Key Components
Binomial Coefficients \(\left(\binom{n}{k}\right.\) or \(\left.{ }^n C_k\right)\) : These numbers (like \(1,4,6,4,1\) for \(n=4\) ) are found in Pascal’s triangle and calculated as \(n!/(k!(n-k)!)\).
Exponents: The power of ‘ \(a\) ‘ starts at \(n\) and decreases to 0, while the power of ‘ \(b\) ‘ starts at 0 and increases to \(n\) in each term.
Terms: There are always \(n+1\) terms in the expansion.
For Example: \((x+y)^3\)
Using the theorem: \({ }^3 C_0 x^3 y^0+{ }^3 C_1 x^2 y^1+{ }^3 C_2 x^1 y^2+{ }^3 C_3 x^0 y^3\).
Simplified: \(1 x^3+3 x^2 y+3 x y^2+1 y^3\).
Thus in the expansion of \((x+y)^7\), the coefficient of \(x^3 y^4\) is equivalent to number of ways \(x, x, x, y, y, y, y\) can be arranged which is \(7!/(3!4!)={ }^7 C_3\).
Hence, in general
\(
(x+y)^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1} y+{ }^n C_2 x^{n-2} y^2+\cdots+{ }^n C_n y^n, \quad \text { Where } n \in N
\)
Remark: The above expansion is also valid when \(x\) and \(y\) are complex numbers.
Properties of Binomial Expansion
Property 1: We have,
\(
(x+y)^n=\sum_{r=0}^n{ }^n C_r x^{n-r} y^r
\)
Since \(r\) can have values from 0 to \(n\), the total number of terms in the expansion is \((n+1)\).
For an example of the binomial expansion for \(n=3\) is \((\mathbf{x}+\mathbf{y})^{\mathbf{3}}=\mathbf{x}^{\mathbf{3}}+\mathbf{3 x}^{\mathbf{2}} \mathbf{y}+\mathbf{3 x y}^{\mathbf{2}}+\mathbf{y}^{\mathbf{3}}\), which has 4 (3+1) terms.
Property-2: This property is a key characteristic of the Binomial Theorem expansion of the expression \((x+y)^n\). The sum of the exponents (indices) for variables \(x\) and \(y\) within every single term of the expansion is indeed \(n\).
Explanation:
When expanding a binomial raised to a power \(n\), such as \((x+y)^n\), each term in the resulting series follows this rule. The general form of the expansion is:
\(
(x+y)^n=\sum_{k=0}^n\binom{n}{k} x^{n-k} y^k
\)
In the general term \(\binom{n}{k} x^{n-k} y^k\), the exponent of \(x\) is \((n-k)\) and the exponent of \(y\) is \(k\). Their sum is \((n-k)+k=n\).
For example: Expanding \((x+a)^3\)
Consider the expansion of \((x+a)^3\) :
\(
(x+a)^3=1 x^3 a^0+3 x^2 a^1+3 x^1 a^2+1 x^0 a^3
\)
Observe the indices in each term:
Term 1: \(3+0=3\)
Term 2: \(2+1=3\)
Term 3: \(1+2=3\)
Term 4: \(0+3=3\)
In every term, the sum of the indices of \(x\) and \(y\) equals the power \(n=3\).
Property 3: The coefficients of terms equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficients. This symmetry is a key characteristic of binomial coefficients and is visually represented in Pascal’s Triangle, where each row reads the same forwards and backwards.
\(
\begin{aligned}
&\text { We have, }\\
&\begin{aligned}
{ }^n C_r & ={ }^n C_{n-r}, \quad r=0,1,2, \ldots, n \\
\Rightarrow \quad{ }^n C_0 & ={ }^n C_n,{ }^n C_1={ }^n C_{n-1},{ }^n C_2={ }^n C_{n-2}=\ldots
\end{aligned}
\end{aligned}
\)
For example: Expansion of \((x+y)^4\)
Consider the expansion of \((x+y)^4\). The binomial theorem gives:
\(
(x+y)^4={ }^4 C_0 x^4 y^0+{ }^4 C_1 x^3 y^1+{ }^4 C_2 x^2 y^2+{ }^4 C_3 x^1 y^3+{ }^4 C_4 x^0 y^4
\)
Calculate the coefficients:
First term coefficient (from the beginning, \(r=0\) ):
\(
{ }^4 C_0=\frac{4!}{0!(4-0)!}=\frac{4!}{1 \cdot 4!}=1
\)
Last term coefficient (from the end, which corresponds to \(r=n-0=4\) ):
\(
{ }^4 C_4=\frac{4!}{4!(4-4)!}=\frac{4!}{4!\cdot 0!}=1
\)
Second term coefficient (from the beginning, \(r=1\) ):
\(
{ }^4 C_1=\frac{4!}{1!(4-1)!}=\frac{4!}{1!\cdot 3!}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{1 \cdot(3 \cdot 2 \cdot 1)}=4
\)
Second to last term coefficient (from the end, which corresponds to \(r=n-1=3\) ):
\(
{ }^4 C_3=\frac{4!}{3!(4-3)!}=\frac{4!}{3!\cdot 1!}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1) \cdot 1}=4
\)
Middle term coefficient (from the beginning, \(r=2\) ):
\(
{ }^4 C_2=\frac{4!}{2!(4-2)!}=\frac{4!}{2!\cdot 2!}=\frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1) \cdot(2 \cdot 1)}=\frac{24}{4}=6
\)
Observe the symmetry:
\({ }^4 C_0={ }^4 C_4=1\)
\({ }^4 C_1={ }^4 C_3=4\)
The expansion is:
\(
(x+y)^4=1 x^4+4 x^3 y+6 x^2 y^2+4 x y^3+1 y^4
\)
The sequence of coefficients \((1,4,6,4,1)\) clearly shows this symmetry; the first is equal to the last, the second is equal to the second to last, and so on. This is the essence of the property.
Property 4: Thus, the terms in the expansion of \((x-y)^n\) are alternatively positive and negative, the last term is positive or negative according as \(n\) is even or odd.
Replacing \(y\) by \(-y\), in the expansionof \((x+y)^n\), we get
\(
\begin{aligned}
(x-y)^n= & { }^n C_0 x^n y^0-{ }^n C_1 x^{n-1} y^1+{ }^n C_2 x^{n-2} y^2-{ }^n C_3 x^{n-3} y^3 \\
& \ldots \ldots+\ldots+(-1)^{r~} { }^n C_r x^{n-r} y^r+\ldots+(-1)^{n~} { }^n C_n x^0 y^n .
\end{aligned}
\)
i.e., \(\quad(x-y)^n=\sum_{r=0}^n(-1)^r~ { }^n C_r x^{n-r} y^r\)
The terms go from \(r=0\) to \(r=n\).
The first term has \(r=0\) (even), so its sign is positive.
The second term has \(r=1\) (odd), so its sign is negative.
The signs continue to alternate throughout the expansion.
For the last term in the expansion, the value of \(r\) is \(n\).
If \(n\) is even, the last term will have an even exponent for ( \(-y\) ), making it positive.
If \(n\) is odd, the last term will have an odd exponent for ( \(-y\) ), making it negative.
For examples
For \(n=2\) (even): \((x-y)^2=x^2-2 x y+y^2\). The last term, \(y^2\), is positive.
For \(n=3\) (odd): \((x-y)^3=x^3-3 x^2 y+3 x y^2-y^3\). The last term, \(-y^3\), is negative.
For \(n=4\) (even): \((x-y)^4=x^4-4 x^3 y+6 x^2 y^2-4 x y^3+y^4\). The last term, \(y^4\), is positive.
Property-5: Putting \(x=1\) and \(y=x\) in the expansion of \((x+y)^n\), we get
\(
(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots+{ }^n C_r x^r+\ldots+{ }^n C_n x^n
\)
\(
\text { i.e., }(1+x)^n=\sum_{r=0}^n{ }^n C_r x^r
\)
Property-6: Putting \(x=1\) and \(y=-x\) in the expansion of \((x+y)^n\), we get
\(
\begin{aligned}
(1-x)^n={ }^n C_0-{ }^n C_1 x & +{ }^n C_2 x^2-{ }^n C_3 x^3+\ldots \\
& +(-1)^r~{ }^n C_r x^r+\ldots+(-1)^n~{ }^n C_n x^n
\end{aligned}
\)
i.e., \(\quad(1-x)^n=\sum_{r=0}^n(-1)^r~ { }^n C_r x^r\)
Example 1: If in the expansion of \((1+x)^m(1-x)^n\), the coefficients of \(x\) and \(x^2\) are 3 and -6 respectively, then \(m\) is [IIT 1999, JEE (WB) 2007]
(a) 6
(b) 9
(c) 12
(d) 24
Solution: We have,
\((1+x)^m(1-x)^n\)
\(
=\left({ }^m C_0+{ }^m C_1 x+{ }^m C_2 x^2+\ldots .+{ }^m C_m x^m\right)\times\left({ }^n C_0-{ }^n C_1 x+{ }^n C_2 x^2 \ldots . .+(-1)^{n n} C_n x^n\right)
\)
\(
={ }^m C_0 \cdot{ }^n C_0-\left({ }^m C_0{ }^n C_1-{ }^n C_0{ }^m C_1\right) x+\left({ }^m C_0{ }^n C_2+{ }^n C_0{ }^m C_2-{ }^m C_1 { }^n C_1\right) x^2+\ldots \ldots
\)
It is given that the coefficients of \(x\) and \(x^2\) in the expansion of \((1+x)^m(1-x)^n\) are 3 and -6 respectively.
\(
\begin{aligned}
& \therefore \quad-\left({ }^m C_0{ }^n C_1-{ }^n C_0{ }^m C_1\right)=3 \\
& \text { and, } \quad{ }^m C_0{ }^n C_2+{ }^n C_0{ }^m C_2-{ }^m C_1{ }^n C_1=-6
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & m-n=3 \text { and } n(n-1)+m(m-1)-2 m n=-12 \\
\Rightarrow & m-n=3 \text { and }(m-n)^2-(m+n)=-12 \\
\Rightarrow & m-n=3 \text { and } m+n=21 \\
\Rightarrow & m=12, n=9
\end{array}
\)
Property-7: The coefficient of \((r+1)^{\text {th }}\) term in the expansion of \((1+x)^n\) is \({ }^n C_r\).
Example 2: Find the coefficient of the fifth term in the expansion of \((1+x)^7\)
Solution: To find the coefficient of the fifth term in the expansion of \((1+x)^7\), we set \(n=7\) and \(r+1=5\), which means \(r=4\). The coefficient is:
\(
{ }^7 C_4=\frac{7!}{4!(7-4)!}=\frac{7 \times 6 \times 5 \times 4!}{4!\times 3!}=\frac{210}{6}=35
\)
The fifth term of the expansion is \(35 x^4\).
Property-8: The coefficient of \(x^r\) in the expansion of \((1+x)^n\) is \({ }^n C_r\).
Example 3: Find the coefficient of \(x^2\) in the expansion of \((1+x)^4\) is 6.
Solution: Identify \(\boldsymbol{n}\) and \(\boldsymbol{r}\)
For the general expansion of \((1+x)^n\), we want to find the coefficient of the term \(x^r\). In this example, we will find the coefficient of \(x^2\) in the expansion of \((1+x)^4\).
Comparing the expression to the general form, we identify \(n=4\) and \(r=2\).
The coefficient of \(x^r\) in \((1+x)^n\) is given by the formula \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\).
\(
=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)}=\frac{24}{4}=6
\)
The coefficient of \(x^2\) in the expansion of \((1+x)^4\) is 6.
Example 4: The coefficient of \(x^{20}\) in \(\left(1+3 x+3 x^2+x^3\right)^{20}\), is
(a) \({ }^{60} \mathrm{C}_{40}\)
(b) \({ }^{30} \mathrm{C}_{20}\)
(c) \({ }^{15} \mathrm{C}_2\)
(d) none of these
Solution: (a) We have,
\(
\begin{array}{ll}
\therefore & \left(1+3 x+3 x^2+x^3\right)^{20}=\left\{(1+x)^3\right\}^{20}=(1+x)^{60} \\
\therefore & \text { Coefficient of } x^{20} \text { in }\left(1+3 x+3 x^2+x^3\right)^{20} \\
& =\text { Coefficient of } x^{20} \text { in }(1+x)^{60}={ }^{60} C_{20}={ }^{60} C_{40}
\end{array}
\)
Example 5: The coefficient of \(x^{100}\) in the expansion of \(\sum_{r=0}^{200}(1+x)^r\) is [CEE (Delhi) 2005]
(a) \(\binom{200}{100}\)
(b) \(\binom{201}{102}\)
(c) \(\binom{200}{101}\)
(d) \(\binom{201}{101}\)
Solution: (d) We have,
Step 1: Sum the geometric series
The given expression is a geometric series with the first term \(a=1\), common ratio \(r=(1+x)\), and number of terms \(n=201\). The sum \(S\) is calculated using the formula \(S_n=a \frac{r^n-1}{r-1}:\)
\(
S=\frac{(1+x)^{201}-1}{(1+x)-1}=\frac{(1+x)^{201}-1}{x}
\)
We can expand the term \((1+x)^{201}\) using the binomial theorem \(\left(\sum_{k=0}^n\binom{n}{k} x^k\right)\) and simplify the entire expression:
\(
\begin{gathered}
S=\frac{\left(1+\binom{201}{1} x+\binom{201}{2} x^2+\ldots+\binom{201}{201} x^{201}\right)-1}{x} \\
S=\frac{\binom{201}{1} x+\binom{201}{2} x^2+\ldots+\binom{201}{201} x^{201}}{x} \\
S=\binom{201}{1}+\binom{201}{2} x+\ldots+\binom{201}{201} x^{200}
\end{gathered}
\)
Step 2: Find the coefficient of \(\boldsymbol{x}^{\mathbf{1 0 0}}\)
The terms in the expanded series \(S\) have the form \(\binom{201}{k} x^{k-1}\). To find the coefficient of \(x^{100}\), we set the exponent to 100 : \(k-1=100\), which gives \(k=101\).
The corresponding coefficient is \(\binom{201}{101}\).
Note: Using the formula for the sum of a geometric series \(S_n=a \frac{\left(r^n-1\right)}{r-1}\) where \(a=1\), \(r=(1+x)\), and \(n=201\) terms:
\(
\sum_{r=0}^{200}(1+x)^r=\frac{(1+x)^{201}-1}{(1+x)-1}=\frac{(1+x)^{201}-1}{x}=\frac{1}{x}\left((1+x)^{201}-1\right)
\)
General Term In Binomial Expansion
We have,
\(
\begin{aligned}
& (x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 x^{n-2} a^2+\ldots \\
& +\ldots .+{ }^n C_r x^{n-r} a^r+\ldots+{ }^n C_n x^0 a^n
\end{aligned}
\)
We observe that the \((r+1)^{\text {th }}\) term is given by
\(T_{r+1}={ }^n C_r x^{n-r} a^r\)
This is called the general term, because by giving different values to \(r\) we can determine all terms of the expansion.
Remark:
In the binomial expansion of \((x-a)^n\), the general term is given by
\(
T_{r+1}=(-1)^r~{ }^n C_r x^{n-r} a^r
\)
For example, Find the 4th term in the expansion of \((x-2)^5\).
Identify parameters: Here, \(x\) is the first term, \(a=2\), and \(n=5\). To find the 4 th term, we use \(r+1=4\), so \(r=3\).
Substitute into the formula:
\(
T_4=T_{3+1}=(-1)^{35} C_3 x^{5-3}(2)^3
\)
Calculate the coefficient and simplify:
\(
\begin{aligned}
& { }^5 C_3=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=\frac{5 \times 4}{2 \times 1}=10 \\
& T_4=(-1) \times 10 \times x^2 \times 8=-80 x^2
\end{aligned}
\)
The 4th term is \(-80 x^2\).
In the binomial expansion of \((1+x)^{n}\), we have
\(
T_{r+1}={ }^n C_r x^{r}
\)
For example, Find the 3rd term in the expansion of \((1+x)^4\).
Identify parameters: Here, \(n=4\). For the 3rd term, \(r+1=3\), so \(r=2\).
Substitute into the formula:
\(
T_3=T_{2+1}={ }^4 C_2 x^2
\)
Calculate the coefficient:
\(
{ }^4 C_2=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4 \times 3}{2 \times 1}=6
\)
The 3rd term is \(6 x^2\).
In the binomial expansion of \((1-x)^n\), we have
\(
T_{r+1}=(-1)^r~{ }^n C_r x^r
\)
For example, Find the 5th term in the expansion of \((1-x)^6\).
Identify parameters: Here, \(n=6\). For the 5th term, \(r+1=5\), so \(r=4\).
Substitute into the formula:
\(
T_5=T_{4+1}=(-1)^{46} C_4 x^4
\)
Calculate the coefficient:
\(
{ }^6 C_4={ }^6 C_{6-4}={ }^6 C_2=\frac{6!}{2!(6-2)!}=\frac{6 \times 5}{2 \times 1}=15
\)
The 5th term is \(15 x^4\).
In the binomial expansion of \((x+a)^n\), the \(r\) th term from the end is \(\{(n+1)-r+1\}=(n-r+2)^{\text {th }}\) term from the beginning.
For example, find the 3rd term from the end in the expansion of \((x+a)^5\).
Identify parameters: Here, \(n=5\) and we want the \(r=3^{\text {rd }}\) term from the end.
Calculate its position from the beginning: The position is \((5-3+2)^{\text {th }}\) term, which is the \(4^{\text {th }}\) term from the beginning.
Use the general term formula for the beginning: For the 4th term, \(R+1=4\), so \(R=3\). The general term is \(T_{R+1}={ }^n C_R x^{n-R} a^R\).
\(
T_4={ }^5 C_3 x^{5-3} a^3
\)
Calculate and simplify:
\(
\begin{aligned}
& { }^5 C_3=10 \\
& T_4=10 x^2 a^3
\end{aligned}
\)
The 3rd term from the end is \(10 x^2 a^3\). This corresponds to the symmetry of binomial coefficients, where the term from the end can also be calculated as
\(
{ }^n C_{n-r+1} x^{r-1} a^{n-r+1} .
\)
Example 6: Find the 3-rd term in the expansion of \((x+2)^{10}\)
Solution: Here, \(n=10\), the first term is \(x\), the second term is 2, and for the 3rd term, \(r=2\) (as the term number is \(r+1\)).
Using the formula:
\(
T_3=T_{2+1}={ }^{10} C_2 x^{10-2}(2)^2
\)
Calculate the combination and simplify:
\(
\begin{aligned}
& { }^{10} C_2=\frac{10!}{2!(10-2)!}=\frac{10 \times 9}{2 \times 1}=45 \\
& T_3=45 \times x^8 \times 4=180 x^8
\end{aligned}
\)
Example 7: Find the term independent of \(x\) in the expansion of \(\left(x+\frac{1}{x^2}\right)^6\)
Solution: Here, \(n=6\), the first term is \(x\), and the second term is \(\frac{1}{x^2}=x^{-2}\).
The general term is:
\(
T_{r+1}={ }^6 C_r x^{6-r}\left(x^{-2}\right)^r={ }^6 C_r x^{6-r} x^{-2 r}={ }^6 C_r x^{6-3 r}
\)
For the term to be independent of \(x\) (a constant term), the exponent of \(x\) must be zero:
\(
6-3 r=0 \Longrightarrow 3 r=6 \Longrightarrow r=2
\)
Substitute \(\boldsymbol{r}=2\) back into the general term formula to find the term:
\(
T_3={ }^6 C_2 x^{6-3(2)}={ }^6 C_2 x^0=\frac{6!}{2!(4)!} \times 1=\frac{6 \times 5}{2 \times 1}=15
\)
The term independent of \(\boldsymbol{x}\) is 15.
Example 8: In the binomial expansion of \((1+a)^{m+n}\), if the coefficients of \(a^m\) and \(a^n\) are \(A\) and \(B\) respectively, then
(a) \(A=B\)
(b) \(m A=n B\)
(c) \(n A=m B\)
(d) \(A=2 B\)
Solution: (a) Step 1: Identify the coefficients of \(a^m\) and \(a^n\)
The general term in the binomial expansion of \((1+a)^{m+n}\) is given by
\(
T_{r+1}=\binom{m+n}{r} a^r .
\)
The coefficient of \(a^m\) occurs when \(r=m\) :
\(
A=\text { Coefficient of } a^m=\binom{m+n}{m}=\frac{(m+n)!}{m!n!}
\)
The coefficient of \(a^n\) occurs when \(r=n\) :
\(
B=\text { Coefficient of } a^n=\binom{m+n}{n}=\frac{(m+n)!}{n!m!}
\)
Step 2: Compare the coefficients
By comparing the expressions for \(\boldsymbol{A}\) and \(\boldsymbol{B}\), we can see they are identical:
\(
\begin{aligned}
& A=\frac{(m+n)!}{m!n!} \\
& B=\frac{(m+n)!}{m!n!}
\end{aligned}
\)
Example 9: In the expansion of \(\left(x^3-\frac{1}{x^2}\right)^{15}\), the constant term, is
(a) \({ }^{15} \mathrm{C}_9\)
(b) 0
(c) \(-{ }^{15} C_9\)
(d) 1
Solution: (c) Let \((r+1)^{\text {th }}\) term be the constant term in the expansion of \(\left(x^3-\frac{1}{x^2}\right)^{15}\).
\(
\begin{array}{ll}
\therefore & T_{r+1}={ }^{15} C_r\left(x^3\right)^{15-r}\left(-\frac{1}{x^2}\right)^r \text { is independent of } x \\
\Rightarrow & T_{r+1}={ }^{15} C_r x^{45-5 r}(-1)^r \text { is independent of } x \\
\Rightarrow & 45-5 r=0 \Rightarrow r=9
\end{array}
\)
Thus, tenth term is independent of \(x\) and is given by
\(
T_{10}={ }^{15} C_9(-1)^9=-{ }^{15} C_9
\)
Example 10: The coefficient of \(x^4\) in the expansion of \(\left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}\), is [CEE (Delhi) 2000]
(a) \(\frac{405}{256}\)
(b) \(\frac{504}{259}\)
(c) \(\frac{450}{263}\)
(d) none of thesse
Solution: (a) Suppose \(x^4\) occurs in \((r+1)^{\text {th }}\) term.
We have,
\(
T_{r+1}={ }^{10} C_r\left(\frac{x}{2}\right)^{10-r}\left(-\frac{3}{x^2}\right)^r={ }^{10} C_r x^{10-3 r}(-3)^r 2^{r-10}
\)
This will contain \(x^4\), if
\(
\therefore \quad 10-3 r=4 \Rightarrow 3 r=6 \Rightarrow r=2
\)
So, \(x^4\) occurs in 3rd term and its coefficient is
\(
{ }^{10} C_2 \times(-3)^2 \times 2^{2-10}={ }^{10} C_2 \times \frac{3^2}{2^8}=\frac{5 \times 9 \times 3^2}{2^8}=\frac{405}{256}
\)
Example 11: The third term in the expansion of \(\left(\frac{1}{x}+x^{\log _{10} x}\right)^5, x>1\) is 1000 , then \(x\) equals
(a) 100
(b) 10
(c) 1
(d) \(1 / \sqrt{10}\)
Solution: (a) We have,
\(
\begin{aligned}
& T_3=1000 \\
\Rightarrow & { }^5 C_2\left(\frac{1}{x}\right)^{5-2}\left(x^{\log _{10} x}\right)^2=1000 \\
\Rightarrow & x^{2 \log _{10}^x-3}=100 \\
\Rightarrow & 2 \log _{10} x-3=\log _x 10^2 \\
\Rightarrow & 2 y-3=\frac{2}{y}, \text { where } y=\log _{10} x \\
\Rightarrow & 2 y^2-3 y-2=0 \\
\Rightarrow & (2 y+1)(y-2)=0 \\
\Rightarrow & y=2 \text { or, } y=1 / 2
\end{aligned}
\)
Now, \(y=2 \Rightarrow \log _{10} x=2 \Rightarrow x=10^2=100\)
and, \(y=-1 / 2 \Rightarrow \log _{10} x=-1 / 2 \Rightarrow x=10^{-1 / 2}\)
But, \(x>1\). So, \(x=100\)
Example 12: If the \(21{ }^{\text {st }}\) and \(22{ }^{\text {nd }}\) terms in the expansion of \((1-x)^{44}\) are equal, then find the value of \(x\)
Solution: \(T_{22}=T_{21} \Rightarrow{ }^{44} C_{21}(-x)^{21}={ }^{44} C_{20}(-x)^{20}\)
\(
\therefore \quad \frac{{ }^{44} C_{21}}{{ }^{44} C_{20}}=\frac{1}{-x} \text { or } \frac{{ }^n C_r}{{ }^n C_{r-1}}=-\frac{1}{x}=\frac{n-r+1}{r}
\)
Put \(n=44, r=21\)
\(
\begin{aligned}
& \therefore \quad-\frac{1}{x}=\frac{44-21+1}{21}=\frac{24}{21}=\frac{8}{7} \\
& \therefore \quad x=-7 / 8
\end{aligned}
\)
Example 13: Find the term in \(\left(\sqrt[3]{\frac{a}{\sqrt{b}}}+\sqrt{\frac{b}{\sqrt[3]{a}}}\right)^{21}\) which has the same power of \(a\) and \(b\).
Solution: The general term in the binomial expansion of \((X+Y)^n\) is \(T_{r+1}={ }^n C_r X^{n-r} Y^r\). Here, \(n=21, X=\sqrt[3]{\frac{a}{\sqrt{b}}}=a^{1 / 3} b^{-1 / 6}\), and \(Y=\sqrt{\frac{b}{\sqrt[3]{a}}}=b^{1 / 2} a^{-1 / 6}\).
\(
\begin{aligned}
&\text { Substituting these into the general term formula: }\\
&\begin{aligned}
T_{r+1} & ={ }^{21} C_r\left(a^{1 / 3} b^{-1 / 6}\right)^{21-r}\left(b^{1 / 2} a^{-1 / 6}\right)^r \\
& ={ }^{21} C_r a^{\frac{1}{3}(21-r)} b^{-\frac{1}{6}(21-r)} a^{-\frac{1}{6} r} b^{\frac{1}{2} r} \\
& ={ }^{21} C_r a^{\left(7-\frac{r}{3}\right)-\frac{r}{6}} b^{\left(-\frac{21}{6}+\frac{r}{6}\right)+\frac{r}{2}} \\
& ={ }^{21} C_r a^{7-\frac{r}{3}-\frac{r}{6}} b^{-\frac{7}{2}+\frac{r}{6}+\frac{r}{2}} \\
& ={ }^{21} C_r a^{7-\frac{3 r}{6}} b^{-\frac{7}{2}+\frac{4 r}{6}} \\
& ={ }^{21} C_r a^{7-\frac{r}{2}} b^{\frac{2 r}{3}-\frac{7}{2}}
\end{aligned}
\end{aligned}
\)
Since the powers of \(a\) and \(b\) are the same, therefore
\(
7-\frac{r}{2}=\frac{2}{3} r-\frac{7}{2} \Rightarrow r=9
\)
Example 14: Find the coefficient of \(x^{50}\) in \((1+x)^{101} \times\left(1-x+x^2\right)^{100}\).
Solution:
\(
\begin{aligned}
(1+x)^{101}\left(1-x+x^2\right)^{100} & =(1+x)\left[(1+x)^{100}\left(1-x+x^2\right)^{100}\right] \\
& =(1+x)\left(1-x^3\right)^{100} \quad\left(\text { since }(1+x)\left(1-x+x^2\right)=1-x^3\right) \\
& =\left(1-x^3\right)^{100}+x\left(1-x^3\right)^{100} .
\end{aligned}
\)
Now consider the powers of \(x\) :
\(\left(1-x^3\right)^{100}\) contains only terms of the form \(x^{3 k}\).
\(x\left(1-x^3\right)^{100}\) contains only terms of the form \(x^{3 k+1}\).
To get an \(x^{50}\) term:
From \(\left(1-x^3\right)^{100}\) : requires \(50=3 k\), impossible.
From \(x\left(1-x^3\right)^{100}\) : requires \(49=3 k\), impossible.
Since neither 50 nor 49 is a multiple of 3 , no such term exists.
Example 15: If sum of the coefficients of the first, second and third terms of the expansion of \(\left(x^2+\frac{1}{x}\right)^m\) is 46, then find the coefficient of the term that does not contain \(x\)
Solution: The general term of
\(
\left(x^2+\frac{1}{x}\right)^m
\)
is
\(
T_{r+1}=\binom{m}{r}\left(x^2\right)^{m-r}\left(\frac{1}{x}\right)^r=\binom{m}{r} x^{2 m-3 r} .
\)
First term ( \(r=0\) ):
\(
\text { Coefficient }=\binom{m}{0}=1
\)
Second term \((r=1)\) :
\(
\text { Coefficient }=\binom{m}{1}=m
\)
Third term \((r=2)\) :
\(
\text { Coefficient }=\binom{m}{2}=\frac{m(m-1)}{2}
\)
Given:
\(
1+m+\frac{m(m-1)}{2}=46
\)
\(
\begin{aligned}
1+m+\frac{m^2-m}{2} & =46 \\
2+2 m+m^2-m & =92 \\
m^2+m-90 & =0 \\
(m-9)(m+10) & =0 \\
m=9 \quad(\text { reject } m & =-10)
\end{aligned}
\)
Find the term independent of \(x\)
The power of \(x\) in the general term is:
\(
2 m-3 r
\)
For the term independent of \(x\) :
\(
2 m-3 r=0
\)
Substitute \(m=9\) :
\(
18-3 r=0 \Rightarrow r=6
\)
The term independent of \(x\) is the 7 th term (since \(r=6\) ), and its coefficient is \({ }^9 C_6\).
\(
{ }^9 C_6=\frac{9!}{6!3!}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=3 \times 4 \times 7=84
\)
Example 16: Find the number of terms which are free from radical signs in the expansion of \(\left(y^{1 / 5}+x^{1 / 10}\right)^{55}\).
Solution: In the expansion of \(\left(y^{1 / 5}+x^{1 / 10}\right)^{55}\),
\(
T_{r+1}={ }^{55} C_r\left(y^{1 / 5}\right)^{5.5-r}\left(x^{1 / 10}\right)^r={ }^{55} C_r y^{11-r / 5} x^{r / 10}
\)
Thus \(T_{r+1}\) will be independent of radicals if the exponents \(r / 5\) and \(r / 10\) are integers for \(0 \leq r \leq 55\), which is possible only when \(r=0,10,20,30,40,50\).
Therefore, there are six terms, i.e., \(T_1, T_{11}, T_{21}, T_{31}, T_{41}, T_{51}\) which are independent of radicals.
Example 17: If the second term in the expansion of \(\left(\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}}\right)^n\) is \(14 a^{5 / 2}\), then the value of \(\frac{{ }^n C_3}{{ }^n C_2}\), is
(a) 4
(b) 3
(c) 12
(d) 6
Solution: (a) We have,
The binomial expression is \(\left(\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}}\right)^n\).
The terms can be simplified as \(x=a^{1 / 13}\) and \(y=\frac{a}{a^{-1 / 2}}=a^{1-(-1 / 2)}=a^{3 / 2}\).
The expansion is of \(\left(a^{1 / 13}+a^{3 / 2}\right)^n\).
The general term of the expansion of \((x+y)^n\) is \(T_{r+1}=\binom{n}{r} x^{n-r} y^r\). For the second term, \(T_2\), we use \(r=1\) :
\(
\begin{gathered}
T_2=\binom{n}{1}\left(a^{1 / 13}\right)^{n-1}\left(a^{3 / 2}\right)^1 \\
T_2=n \cdot a^{\frac{n-1}{13}} \cdot a^{\frac{3}{2}}=n \cdot a^{\frac{2(n-1)+39}{26}}=n \cdot a^{\frac{2 n+37}{26}}
\end{gathered}
\)
Step 3: Solve for \(\mathbf{n}\)
We are given that the second term \(T_2=14 a^{5 / 2}\).
By comparing coefficients and exponents with our expression for \(T_2\) :
\(
n \cdot a^{\frac{2 n+37}{26}}=14 a^{5 / 2}
\)
Comparing coefficients gives \(n=14\). Comparing exponents confirms this:
\(
\frac{2 n+37}{26}=\frac{2(14)+37}{26}=\frac{65}{26}=\frac{5}{2} .
\)
Calculate the required ratio
We need to find the value of \(\frac{{ }^n C_3}{{ }^n C_2}\) with \(n=14\).
Using the formula \(\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n-r+1}{r}\) for \(r=3\) :
\(
\frac{{ }^{14} C_3}{{ }^{14} C_2}=\frac{14-3+1}{3}=\frac{12}{3}=4
\)
Tips for finding coefficients of required term
Example 18: Find the \(6^{\text {th }}\) term in the expansion of \(\left(\mathbf{2} \boldsymbol{x}^{\mathbf{2}}-\mathbf{1} / \mathbf{3} \boldsymbol{x}^{\mathbf{2}}\right)^{\mathbf{1 0}}\).
Solution: For \((a+b)^n\),
\(
T_{r+1}=\binom{n}{r} a^{n-r} b^r={ }^n C_r a^{n-r} b^r
\)
Here:
\(a=2 x^2\)
\(b=-\frac{1}{3 x^2}\)
\(n=10\)
The 6th term corresponds to:
\(
r+1=6 \Rightarrow r=5
\)
Compute the 6th term
\(
\begin{aligned}
T_6 & =\binom{10}{5}\left(2 x^2\right)^5\left(-\frac{1}{3 x^2}\right)^5 \\
& =\binom{10}{5} \cdot 2^5 \cdot(-1)^5 \cdot \frac{x^{10}}{3^5 x^{10}}
\end{aligned}
\)
Notice:
\(
x^{10} \text { cancels out }
\)
Numerical simplification
\(
\begin{gathered}
\binom{10}{5}=252, \quad 2^5=32, \quad 3^5=243 \\
T_6=-\frac{252 \times 32}{243} \\
=-\frac{8064}{243}=-\frac{896}{27}
\end{gathered}
\)
Example 19: Find the coefficient of \(x^5\) in the expansion of \(\left(1+x^2\right)^5 .(1+x)^4\) is 60.
Solution: Coefficient of \(x^5\) in \(\left(1+x^2\right)^5 \cdot(1+x)^4\)
\(\begin{array}{r}=\text { Coefficient of } x^5 \text { in }\left({ }^5 C_0+{ }^5 C_1 x^2+{ }^5 C_2 x^4+{ }^5 C_3 x^6+\cdots\right) \\ \left({ }^4 C_0+{ }^4 C_1 x+{ }^4 C_2 x^2+\cdots\right)\end{array}\)
\(
\begin{aligned}
& ={ }^5 C_1{ }^4 C_3+{ }^5 C_2{ }^4 C_1 \\
& =20+40 \\
& =60
\end{aligned}
\)
Explanation: Step 1: Write the relevant expansions
\(
\begin{gathered}
\left(1+x^2\right)^5=\binom{5}{0}+\binom{5}{1} x^2+\binom{5}{2} x^4+\binom{5}{3} x^6+\cdots \\
(1+x)^4=\binom{4}{0}+\binom{4}{1} x+\binom{4}{2} x^2+\binom{4}{3} x^3+\cdots
\end{gathered}
\)
Step 2: Identify terms whose powers add to \(x^5\)
We need:
\(
x^{2 k} \cdot x^m=x^5 \quad \Rightarrow \quad 2 k+m=5
\)
Valid combinations:
\(x^2 \cdot x^3\)
\(x^4 \cdot x^1\)
Step 3: Add their coefficients
\(
\begin{gathered}
\binom{5}{1}\binom{4}{3}+\binom{5}{2}\binom{4}{1} \\
=5 \cdot 4+10 \cdot 4 \\
=20+40=60
\end{gathered}
\)
A useful special case of the Binomial Theorem is
\(
(1+x)^n=\sum_{k=0}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) x^k
\)
for any positive integer \(n\), which is just the Taylor series for \((1+x)^n\).
This formula can be extended to all real powers \(\alpha\) :
\(
(1+x)^\alpha=\sum_{k=0}^{\infty}\left(\begin{array}{l}
\alpha \\
k
\end{array}\right) x^k
\)
for any real number \(\boldsymbol{\alpha}\), where
\(
\left(\begin{array}{l}
\alpha \\
k
\end{array}\right)=\frac{(\alpha)(\alpha-1)(\alpha-2) \cdots(\alpha-(k-1))}{k !}=\frac{\alpha !}{k !(\alpha-k) !}
\)
Notice that the formula now gives an infinite series: when \(\alpha=n\) is a positive integer, all but the first \((n+1)\) terms are 0 since after this \(
(n-n) ! = 0 !\) appears in each numerator.
This expansion is very useful for approximating \((1+x)^\alpha\) for \(|x| \ll 1\) :
\(
(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3 !} x^3+\cdots
\)
But for \(|x| \ll 1\), higher powers of \(x\) get small very quickly, so \((1+x)^\alpha\) can be approximated to any accuracy we need by truncating the series after a finite number of terms.
For example, \(|x| \ll 1\),
\(
\begin{aligned}
(1+x)^{5 / 2} & \approx 1+\frac{5}{2} x \\
(1-2 x)^{100} & \approx 1-200 x \\
\left(1+x^2\right)^{-3} & \approx 1-3 x^2
\end{aligned}
\)
This type of reasoning is useful in investigating what happens when a physical system is perturbed slightly, introducing a new very small term \(\boldsymbol{x}\).
Example 20: If \(x\) is so small that \(x^3\) and higher powers of \(x\) may be neglected, then \(\frac{(1+x)^{3 / 2}-\left(1+\frac{x}{2}\right)^3}{(1-x)^{1 / 2}}\) may be approximated as [AIEEE 2005]
(a) \(\frac{x}{2}-\frac{3}{8} x^2\)
(b) \(-\frac{3}{8} x^2\)
(c) \(3 x+\frac{3}{8} x^2\)
(d) \(1-\frac{3}{8} x^2\)
Solution: (b) We have,
\(
(1+x)^n \simeq 1+n x+\frac{n(n-1)}{2} x^2 \text { [Given] }
\)
\(
\begin{array}{ll}
\therefore & \frac{(1+x)^{3 / 2}-\left(1+\frac{x}{2}\right)^3}{(1-x)^{1 / 2}} \\
& =\left\{\left(1+\frac{3}{2} x+\frac{3}{8} x^2\right)-\left(1+\frac{3}{2} x+\frac{3}{4} x^2\right)\right\}\left\{1+\frac{1}{2} x+\frac{3}{8} x^2\right\} \\
& =-\frac{3}{8} x^2\left(1+\frac{x}{2}+\frac{3}{8} x^2\right) \simeq-\frac{3}{8} x^2
\end{array}
\)
Explanation:
\(
(1+x)^n=\sum_{k=0}^{\infty}\binom{n}{k} x^k=\binom{n}{0} x^0+\binom{n}{1} x^1+\binom{n}{2} x^2+\binom{n}{3} x^3+\ldots
\)
where the generalized binomial coefficient is defined as
\(
\binom{n}{k}=\frac{n(n-1)(n-2) \ldots(n-k+1)}{k!} .
\)
Calculate the first three binomial coefficients
We calculate the first three coefficients for \(k=0, k=1\), and \(k=2\) :
For \(k=0:\binom{n}{0}=\frac{1}{0!}=1\)
For \(k=1:\binom{n}{1}=\frac{n}{1!}=n\)
For \(k=2:\binom{n}{2}=\frac{n(n-1)}{2!}=\frac{n(n-1)}{2}\)
Substitute coefficients into the expansion
Substituting these coefficients back into the series expansion yields the full expression:
\(
\begin{gathered}
(1+x)^n=1 \cdot x^0+n \cdot x^1+\frac{n(n-1)}{2} \cdot x^2+\frac{n(n-1)(n-2)}{6} \cdot x^3+\ldots \\
(1+x)^n=1+n x+\frac{n(n-1)}{2} x^2+\ldots
\end{gathered}
\)
The approximation \(\simeq 1+n x+\frac{n(n-1)}{2} x^2\) is formed by truncating the infinite series after the second-order \(\left(x^2\right)\) term. This approximation is highly accurate for small values of \(x\) (specifically when \(|x|<1\) ), because the higher-order terms (\(x^3, x^4\), etc.) become very small and negligible, allowing us to approximate the function using just the first three terms.
You cannot copy content of this page