Question

Let sum of the series $2+4+7+11+16+............$ to $n$ terms be $=\frac{1}{m}n(n^2+3n+k).$ Find $k-m$ ?

Answer 1

$S_n=2+4+7+11+\mathrm{16.......}+T_n$

$S_n=0+2+4+7+...+T_{n-1}+T_n$

Subtracting both, we get

$T_n=2+2+3+4+.......$

$T_n-2=(2+3+4+.....)$ ($n-1$ Terms)

$T_n-2=\frac{n-1}{2}(2\times 2+(n-2\left)\right)$

$T_n-2=\frac{(n-1)(n+2)}{2}$
$T_n=\frac{1}{2}(n^2+n+2)$
$T_n=\frac{1}{2}n(n+1)+1$
$S_n=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})+n$
$S_n=\frac{n}{12}\left(\right(n+1\left)\right(2n+1)+3(n+1)+12)$
$S_n=\frac{n}{12}(2n^2+3n+1+3n+3+12)$
$S_n=\frac{n}{12}(2n^2+6n+16)$
$S_n=\frac{n}{6}(n^2+3n+8)$
$m=8$ and $k=6$

So, $k-m=2$