Question

Sum of $\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+...$ to $n$ terms be $S=\frac{1}{k}n(n+1)(m{n}^2+23n+34).$ Find $k-m$ ?

Answer 1

$S=\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+...S_r={\sum }_{r=1}^{n}r(r+1)(r+2)={\sum }_{r=1}^n(r^3+{3r}^2+2r)={\sum }_{r=1}^n{r}^3+3{\sum }_{r=1}^n{r}^2+2{\sum }_{r=1}^{n}r$
$={\left(\frac{n(n+1)}{2}\right)}^2+3\frac{\left(n(n+1)(2n+1)\right)}{6}+2\left(\frac{n(n+1)}{2}\right)$
$=n(n+1)(\frac{n(n+1)}{4}+\frac{(2n+1)}{2}+1)$
$=\frac{1}{4}n(n+1)(n^2+n+4n+2+4)=\frac{1}{4}n(n+1)(n^2+5n+6)\therefore k-m=4-1=3$