Question

Write the sum of 20 terms of the series: $1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+....$

Answer 1

Let the nth term of the given series is $T_n$ and $S_n$ be the sum of the given series. $\therefore T_n=\frac{1}{n}[1+2+\mathrm{3...}n]=\frac{1}{n}\left[\frac{n}{2}\right(2\times 1+(n-1)\times 1\left)\right]=\frac{1}{n}times\frac{n}{2}[2+n-1]=\frac{1}{2}(n+1)\therefore S_n={\sum }_{k-1}^n\frac{1}{2}(k+1)=\frac{1}{2}{\sum }_{k-1}^{n}k+\frac{1}{2}{\sum }_{k-1}^{n}1=\frac{1}{2}\left[\frac{n(n+1)}{2}\right]+\frac{1}{2}\times n=\frac{n(n+1)}{4}+\frac{n}{2}=\frac{n(n+1)+2n}{4}=\frac{n[n+1+2]}{4}=\frac{n[n+3]}{4}Putting,n=20,weget{S}_{20}=\frac{20[20+3]}{4}=5\times 23=115$