Question

Find the $n^{th}$ term and the sum of $n$ term of the series $2+12+36+80+150+252+...$

Answer 1

Let $S=2+12+36+80+150+252+...+T_n$ (i)
$S=2+12+36+80+150+252+...+T_{n-1}+T_n$ (ii)
(i)-(ii) $\Rightarrow$ $T_n=2+10+24+44+70+102+...+(T_n-T_{n-1})$ (iii)
$T_n=2+10+24+44+70+102+...+(T_{n-1}-T_{n-2})+(T_n-T_{n-1})$ (iv)
(iii)-(iv) $\Rightarrow$ $T_n-T_{n-1}=2+8+14+20+26+...$
$=\frac{n}{2}[4+(n-1)6]=n[3n-1]\Rightarrow T_n-T_{n-1}=3n^2-n$
$\therefore$ general term of given series is $\sum (T_n-T_{n-1})=\sum (3n^2-n)=n^3+n^2.$
Hence sum of this series is
$S=\sum n^3+\sum n^2=\frac{n^2{(n+1)}^2}{4}+\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)}{12}(3n^2+7n+2)$
$=\frac{1}{12}n(n+1)(n+2)(3n+1)$