Question

Find the $n^{th}$ term and the sum of n terms of the series $8,16,0,-64,-200,-432,....$

Answer 1

To find the $n^{th}$ term we have to look the succesive difference of the sequence

$8,-16,0,-64,-200,-\mathrm{432.....}8,-16,-64,-136,-\mathrm{232.....}-24,-48,-72,-\mathrm{96....}-24,-24,-24$

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

$u_1=8\Rightarrow A+B+C+D=\mathrm{8.....}\left(i\right)u_2=-16\Rightarrow A+2B+4C+8D=-\mathrm{16.....}\left(ii\right)u_3=0\Rightarrow A+3B+9C+27D=\mathrm{0.....}\left(iii\right)u_4=-64\Rightarrow A+4B+16C+64D=-\mathrm{64....}\left(iv\right)$

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

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