Question

Find the sum of the series s = 1 + $\frac{1}{2}$(1 + 2) + $\frac{1}{3}$(1 + 2 + 3) + $\frac{1}{4}$(1 + 2 + 3 + 4) + .........upto 40 terms.

__

Answer 1

If we want to claculate sum of the series,first find the $n^{th}$ term of the given series and then get the summation of it.

$n^{th}$ term of the given series

$T_n$ = $\frac{{\sum }_{i=1}^{n}r}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$ --------(1)

$S_n$ = $\sum$ $T_n$ = $\frac{1}{2}$ $\sum$ (n+1).....

= $\frac{1}{2}$ $[{\sum }_{i=1}^{n}n+{\sum }_{i=1}^{n}1]$

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

Substitute n=50 in equation 2

$S_{40}$ = $\frac{1}{2}$ $[\frac{40\times 41}{2}+40]$

=430