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

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,
$0=3+(4 + 6 + 8 + 10 + . . . . . . . . + (T_{n} - T_{n - 1})-T_{n}$
$\Rightarrow T_{n}=3+(4 + 6 + 8 + 10 + . . . . . . . . (n - 1) terms)=3+\frac{\left(n - 1\right)}{2}\left[8 + \left(n - 2\right) 2\right]$
$=3+\left(n - 1\right)\left(4 + n - 2\right)$
$=3+\left(n - 1\right)\left(n + 2\right)$
$=3+n^{2}+n-2$
$=n^{2}+n+1$
$S_{n}=\Sigma T_{n}=\Sigma n^{2}+\Sigma n+\Sigma 1$
$=\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}+\frac{n \left(n + 1\right)}{2}+n$
If $n=50,$ $S_{50}=\frac{50 \times 51 \times 101}{6}+\frac{50 \times 51}{2}+50$
$=25\times 17\times 101+25\times 51+50=44250$