Question

Find the sum upto n term of the series
3+7+13+21+31+........ nterms

Answer 1

It is nor an $AP$ or $GP$

Let $S_n=3+7+13+21+31+....+a_{n-1}+a_n$ ----- ( 1 )

$S_n=0+3+7+13+21+...+a_{n-2}+a_{n-1}+a_n$ ----- ( 2 )

Subtracting ( 2 ) from ( 1 )

$S_n-S_n=3-0\left[\right(7-3)+(13-7)+(21-13)+....+(a_{n-1}-a_{n-2})+(a_n-a_{n-1}\left)\right]-a_n$

$0=3+[4+6+8+....+a_{n-1}]-a_n$

$a_n=3+[4+6+8+....+a_{n-1}]$ ----- ( 3 )

Now, $4+6+8+....+a_{n-1}$ is an AP

Whose first term $a=4$

Common difference $d=6-4=2$

We know that,

Sum of $n$ terms of $AP=\frac{n}{2}(2a+(n-1\left)d\right)$

Putting, $n=n-1,a=4,d=2$

$[4+6+8+...(n-1\left)terms\right]=\left(\frac{n-1}{2}\right)[2a+(n-1-1\left)d\right]$

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

$=\left(\frac{n-1}{2}\right)[8+2n-4]$

$=\left(\frac{n-1}{2}\right)(2n+4)$

$=\left(\frac{n-1}{2}\right)2(n+2)$

$=(n-1)(n+2)$

Thus, $[4+6+\mathrm{8...}upto(n-1\left)terms\right]=(n-1)(n+2)$

From ( 3 )

$a_n=3+[4+6+8+...+a_{n-1}]$

Putting values

$=3+(n-1)(n+2)$

$=3+n(n+2)-1(n+2)$

$=3+n^2+2n-n-2$

$=n^2+n+3-2$

$=n^2+n+1$

Now,

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

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

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

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

$=\frac{n(n+1)(2n+1)+3n(n+1)+6n}{6}$

$=n\left(\frac{(n+1)(2n+1)+3(n+1)+6}{6}\right)$

$=n\left(\frac{2n^2+n+2n+1+3n+3+6}{6}\right)$

$=n\left(\frac{2n^2+6n+10}{6}\right)$

$=\frac{n}{6}\times 2(n^2+3n+5)$

$=\frac{n}{3}(n^2+3n+5)$

$\therefore$ The required sum is $\frac{n}{3}(n^2+3n+5)$