Question

sum of the series is $1+3+6+10+15+...$ to n terms $\frac{n(n+m)(n+p)}{k}$.Find $k-m-p$ ?

Answer 1

$S=1+3+6+10+15+.....$

$S=1+3+6+10+......$

$a_n=1+2+3+4+5+.........$

$a_n=\frac{n(n+1)}{2}$
$S=\frac{\sum n^2+\sum n}{2}=\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{2}$
$S=\frac{n(n+1)}{4}(\frac{2n+1}{3}+1)$
$S=\frac{n(n+1)(2n+4)}{12}$
$S=\frac{n(n+1)(n+2)}{6}$
$k=6,m=1,n=2$
So $k-m-n=3$