9.8 Sum to n terms of Special Series

We shall now find the sum of first \(n\) terms of some special series, namely;
(i) \(1+2+3+\ldots \quad+n\) (sum of first \(n\) natural numbers)
(ii) \(1^2+2^2+3^2+\ldots+n^2\) (sum of squares of the first \(n\) natural numbers)
(iii) \(1^3+2^3+3^3+\ldots+n^3\) (sum of cubes of the first \(n\) natural numbers).
Let us take them one by one.
(i) \(S _n=1+2+3+\ldots+n\), then \(S _n=\frac{n(n+1)}{2}\)
(ii) Here \(S _n=1^2+2^2+3^2+\ldots+n^2\)
We consider the identity \(k^3-(k-1)^3=3 k^2-3 k+1\) Putting \(k=1,2 \ldots, n \quad\) successively, we obtain
\(
\begin{aligned}
& 1^3-0^3=3(1)^2-3(1)+1 \\
& 2^3-1^3=3(2)^2-3(2)+1 \\
& 3^3-2^3=3(3)^2-3(3)+1
\end{aligned}
\)
\(\ldots\)
\(\ldots\)
\(\ldots\)
\(
n^3-(n-1)^3=3(n)^2-3(n)+1
\)
Adding both sides, we get
\(
\begin{aligned}
& n^3-0^3=3\left(1^2+2^2+3^2+\ldots+n^2\right)-3(1+2+3+\ldots+n)+n \\
& n^3=3 \sum_{k=1}^n k^2-3 \sum_{k=1}^n k+n
\end{aligned}
\)
By (i), we know that \(\sum_{k=1}^n k=1+2+3+\ldots+n=\frac{n(n+1)}{2}\)
Hence
\(
\begin{aligned}
S _n & =\sum_{k=1}^n k^2=\frac{1}{3}\left[n^3+\frac{3 n(n+1)}{2}-n\right]=\frac{1}{6}\left(2 n^3+3 n^2+n\right) \\
& =\frac{n(n+1)(2 n+1)}{6}
\end{aligned}
\)
(iii) Here \(S _n=1^3+2^3+\ldots+n^3\)
We consider the identity, \(\quad(k+1)^4-k^4=4 k^3+6 k^2+4 k+1\) Putting \(k=1,2,3 \ldots n\), we get
\(
\begin{aligned}
&\begin{aligned}
& 2^4-1^4=4(1)^3+6(1)^2+4(1)+1 \\
& 3^4-2^4=4(2)^3+6(2)^2+4(2)+1 \\
& 4^4-3^4=4(3)^3+6(3)^2+4(3)+1
\end{aligned}\\
&\ldots\\
&\ldots\\
&\begin{aligned}
& (n-1)^4-(n-2)^4=4(n-2)^3+6(n-2)^2+4(n-2)+1 \\
& n^4-(n-1)^4=4(n-1)^3+6(n-1)^2+4(n-1)+1 \\
& (n+1)^4-n^4=4 n^3+6 n^2+4 n+1
\end{aligned}
\end{aligned}
\)
Adding both sides, we get
\(
\begin{aligned}
& (n+1)^4-1^4=4\left(1^3+2^3+3^3+\ldots+n^3\right)+6\left(1^2+2^2+3^2+\ldots+n^2\right)+ \\
& 4(1+2+3+\ldots+n)+n
\end{aligned}
\)
\(
=4 \sum_{k=1}^n k^3+6 \sum_{k=1}^n k^2+4 \sum_{k=1}^n k+n \dots(1)
\)
From parts (i) and (ii), we know that
\(
\sum_{k=1}^n k=\frac{n(n+1)}{2} \quad \text { and } \sum_{k=1}^n k^2=\frac{n(n+1)(2 n+1)}{6}
\)
Putting these values in equation (1), we obtain
\(
4 \sum_{k=1}^n k^3=n^4+4 n^3+6 n^2+4 n-\frac{6 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}-n
\)
\(
\begin{aligned}
4 S _n & =n^4+4 n^3+6 n^2+4 n-n\left(2 n^2+3 n+1\right)-2 n(n+1)-n \\
& =n^4+2 n^3+n^2 \\
& =n^2(n+1)^2 .
\end{aligned}
\)
Hence,
\(
S _n=\frac{n^2(n+1)^2}{4}=\frac{[n(n+1)]^2}{4}
\)

Example 1: Find the sum to \(n\) terms of the series: \(5+11+19+29+41 \ldots\)

Solution: Let us write
\(
\begin{aligned}
& S _n=5+11+19+29+\ldots+a_{n-1}+a_n \\
& S _n=\quad 5+11+19+\ldots+a_{n-2}+a_{n-1}+a_n
\end{aligned}
\)
On subtraction, we get
\(
\begin{aligned}
0 & =5+[6+8+10+12+\ldots(n-1) \text { terms }]-a_n \\
a_n & =5+\frac{(n-1)[12+(n-2) \times 2]}{2} \\
& =5+(n-1)(n+4)=n^2+3 n+1
\end{aligned}
\)
Hence
\(
\begin{aligned}
S _n & =\sum_{k=1}^n a_k=\sum_{k=1}^n\left(k^2+3 k+1\right)=\sum_{k=1}^n k^2+3 \sum_1^n k+n \\
& =\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n=\frac{n(n+2)(n+4)}{3} .
\end{aligned}
\)

Example 2: Find the sum to \(n\) terms of the series whose \(n^{\text {th }}\) term is \(n(n+3)\).

Solution: Given that \(a_n=n(n+3)=n^2+3 n\)
Thus, the sum to \(n\) terms is given by
\(
\begin{aligned}
S _n & =\sum_{k=1}^n a_k=\sum_{k=1}^n k^2+3 \sum_{k=1}^n k \\
& =\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}=\frac{n(n+1)(n+5)}{3} .
\end{aligned}
\)