Question

The sum of $50$ terms of the series $3+7+13+21+31+43+.....$ is equal to $S_{50}$ , then the value of $\frac{S_{50}}{12500}$ is

Answer 1

Answer: 3.54

Let, $S=3+7+13+21+31+43+.....+T_n$
$S=3+7+13+21+31+.....+T_{n-1}+T_n$
Subtracting both the equations, we get,
$0=3+(4+6+8+10+........+(T_n-T_{n-1})-T_n$
$\Rightarrow T_n=3+(4+6+8+10+........(n-1)terms)=3+\frac{\left(n-1\right)}{2}\left[8+\left(n-2\right)2\right]$
$=3+\left(n-1\right)\left(4+n-2\right)$
$=3+\left(n-1\right)\left(n+2\right)$
$=3+n^2+n-2$
$=n^2+n+1$
$S_n=\Sigma T_n=\Sigma n^2+\Sigma n+\Sigma 1$
$=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}+n$
If $n=50,$ $S_{50}=\frac{50\times 51\times 101}{6}+\frac{50\times 51}{2}+50$
$=25\times 17\times 101+25\times 51+50=44250$