Question

Find the $n^{th}$ term and the sum of n terms of the series $8,26,54,92,140,198,....$

Answer 1

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

$8,26,54,92,\mathrm{140.....}18,28,38,\mathrm{48.....}10,10,\mathrm{10.....}$

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

For $n=1$ $u_1=8$

$\Rightarrow A+B+C=\mathrm{8.....}\left(i\right)$

For $n=2$ $u_2=26$

$\Rightarrow A+2B+4C=\mathrm{26.....}\left(ii\right)$

For $n=3$ $u_3=54$

$\Rightarrow A+3B+9C=\mathrm{54.....}\left(ii\right)$

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

$\Rightarrow A=0,B=3,C=5$

$\Rightarrow u_n=3n+5n^2$

$S_n=\sum u_n\Rightarrow S_n=3\sum n+5\sum n^2\Rightarrow S_n=3\left[\frac{n(n+1)}{2}\right]+5\left[\frac{n(n+1)(2n+1)}{6}\right]\Rightarrow S_n=\frac{n(n+1)(5n+7)}{3}$