Question

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

Answer 1

$2,12,36,80,150,\mathrm{252.....}10,24,44,70,\mathrm{102.....}14,20,26,\mathrm{32.....}6,6,\mathrm{6....}$

Let $u_n=A+Bn+C{n}^2+D{n}^3$

For $n=1$ $u_1=2$

$\Rightarrow A+B+C+D=\mathrm{2.....}\left(i\right)u_2=12\Rightarrow A+2B+4C+8D=\mathrm{12.....}\left(ii\right)u_3=36\Rightarrow A+3B+9C+27D=\mathrm{36.....}\left(iii\right)u_4=80\Rightarrow A+4B+16C+64D=\mathrm{80....}\left(iv\right)$

Solving $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)$

$\Rightarrow A=0,B=0,C=1,D=1\Rightarrow u_n=n^2+n^3{S}_n=\sum u_n\Rightarrow S_n=\sum n^2+\sum n^3\Rightarrow S_n=\left[\frac{n(n+1)(2n+1)}{6}\right]+\left[\frac{n^2{(n+1)}^2}{4}\right]\Rightarrow S_n=\frac{n(n+1)(n+2)(3n+1)}{12}$