Question

Sum of the series $4+6+9+13+18+...$ to $n$ terms be $=\frac{n}{k}(n^2+3n+m)$.Find $m-k$ ?

Answer 1

$S_n=4+6+9+13+\mathrm{18.......}+T_n$

$S_n=0+4+6+9+...+T_{n-1}+T_n$

Subtracting Both we get

$T_n=4+2+3+4+.......$
$T_n-4=(2+3+4+.....)(n-1Terms)$
$T_n-4=\frac{n-1}{2}(2\times 2+(n-2\left)\right)$
$T_n-4=\frac{(n-1)(n+2)}{2}$
$T_n=\frac{1}{2}(n^2+n+6)$
$T_n=\frac{1}{2}n(n+1)+3$
$S_n=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})+3n$
$S_n=\frac{n}{6}(n^2+3n+20)$
$m=20andk=6.So,m-k=14$.