Question

Find the $nth$ term and the sum of $n$ terms of the series $\mathrm{1.2.4}+\mathrm{2.3.5}+\mathrm{3.4.6}+....$

Answer 1

$\mathrm{1.2.4}+\mathrm{2.3.5}+\mathrm{3.4.6}+\dots .$

$\therefore n^{th}term=n(n+1)(n+3)$

$\therefore$ sum $=\sum _{k=1}^{n}k(k+1)(k+3)$

$=\sum _{k=1}^n(k^3+4k^2+3k)$

$=\sum _{k=1}^n{k}^3+4\sum _{k=1}^n{k}^2+3\sum _{k=1}^{n}k$

${\left[\frac{n(n+1)}{2}\right]}^2+\frac{4n(n+1)(2n+1)}{6}+3\frac{n(n+1)}{2}$

$=\frac{n^2{(n+1)}^2}{4}+\frac{2n(n+1)(2n+1)}{3}+\frac{3}{2}n(n+1)$

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

$=n(n+1)\left[\frac{3n^2+3n+16n+8+18}{12}\right]$

$=\frac{n(n+1)+(3n^2+19n+26)}{12}$ Ans