Question

Find the sum of the first n terms of the series:

3 + 7 + 13 + 21 + 31 + ...

Answer 1

Let $a_n$ be the nth term of given series and $S_n$ be the sum of n-terms of the given series.

$\therefore S_n=3+7+13+21+31+\dots \dots +a_{n-1}+a_n\dots \dots \left(1\right)$

Also $=3+7+13+21+\dots \dots +a_{n-2}+a_{n-1}+a_n\dots \dots \left(2\right)$

Subtracting (2) from (1),

$0=3+(4+6+8+10+\dots \dots \text{upto (n-1) terms})-a_n$

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

$\Rightarrow a_n=3+\frac{n-1}{2}[8+2n-4])$

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

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

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

$\therefore S_n={\sum }_{n=1}^k{a}_k=S_n={\sum }_{n=1}^k[k^2+k+1]$

$=(1^2+1+1)+(2^2+2+1)+(3^2+3+1)+\dots \dots +(n^2+n+1)$

$=(1^2+2^2+3^2+\dots \dots +n^2)+(1+2+3+\dots \dots +n)+n$

$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$

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

$=n\left[\frac{2n^2+6n+10}{6}\right]$

$=\frac{n}{3}[n^2+3n+5]$