Question

Sum of the series $1+(1+2)+(1+2+3)+(1+2+3+4)+...$ to n terms be $\frac{1}{m}n(n+1)[\frac{2n+1}{k}+1]$ .Find the value of k+m.

Answer 1

$T_n=(1+2+\mathrm{3.....}n)=\frac{n(n+1)}{2}=\frac{1}{2}(n^2+n)$

Putting $n=1,2,\mathrm{3.......}n$ on adding, we get

$S_n=\frac{1}{2}(\Sigma n^2+\Sigma n)$

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

$S_n=\frac{1}{4}n(n+1)[\frac{2n+1}{3}+1]=\frac{n(n+1)(n+2)}{6}$

$m=4,k=3So,m+k=7$