Question

Find the sum of the first $n$ terms of the series: $3+7+13+21+31+\cdots$

Answer 1

Step $1$: Finding $n^{th}$ term of given series

The given series is $3+7+13+21+31+...$

$S=3+7+13+21+31+....+a_{n-1}+a_n$

$S=3+7+13+21+...+a_{n-2}+a_{n-1}+a_n$

On subtracting both the equations, we get

$S-S$ $=[3+(7+13+21+31+...+a_{n-1}+a_n\left)\right]$
$-\left[\right(3+7+13+21+31+...+a_{n-1})+a_n]$

$S-S$

$=3+\left[\right(7-3)+(13-7)+(21-13)+...+(a_n-a_{n-1}\left)\right]-a_n$

$0=3+[4+6+8+...(n-1\left)\text{terms}\right]-a_n$

$a_n=3+[4+6+8+...(n-1\left)\text{terms}\right]$

$\Rightarrow a_n=3+\left(\frac{n-1}{2}\right)[2\times 4+(n-1-1\left)2\right]$

$\Rightarrow a_n=3+\left(\frac{n-1}{2}\right)[8+(n-2\left)2\right]$

$\Rightarrow a_n=3+(n-1)(n+2)$

$\Rightarrow a_n=3+(n^2+n-2)$

$\Rightarrow a_n=n^2+n+1$

Step $2$ : Finding sum of given series

$S_n={\sum }_{n=1}^n{a}_n={\sum }_{n=1}^n(n^2+n+1)$

${\sum }_{n=1}^n{a}_n={\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^{n}n+{\sum }_{n=1}^{n}1$

${\sum }_{n=1}^n{a}_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$

${\sum }_{n=1}^n{a}_n=n\left[\frac{(n+1)(2n+1)+3(n+1)+6}{6}\right]$

${\sum }_{n=1}^n{a}_n=n\left[\frac{2n^2+n+2n+1+3n+3+6}{6}\right]$

${\sum }_{n=1}^n{a}_n=n\left[\frac{2n^2+6n+10}{6}\right]$

${\sum }_{n=1}^n{a}_n=2n\left[\frac{n^2+3n+5}{6}\right]$

$S_n={\sum }_{n=1}^n{a}_n=\frac{n}{3}[n^2+3n+5]$