Question

Find the sum of n terms of the series.
3+8+22+72+266+1036+.......

Answer 1

$S=3+8+22+72+266+1036+...$ .....$\left(1\right)$

$S=0+3+8+22+72+266+\mathrm{1036...}$ .....$\left(2\right)$

Eqn$\left(1\right)-$Eqn$\left(2\right)$, we get

$0=3+5+14+50+194+770+.....-T_n$ ....$\left(3\right)$

$T_n=3+5+14+50+194+770+....$ ....$\left(4\right)$

$T_n=3+5+14+50+194+770+....$ ....$\left(5\right)$

$0=3+2+9+36+144+576+.....-t_n$ ....$\left(6\right)$

$t_n=3+2+$ G.P. with first term $9$, common ratio $4$, and number of terms are $n-2$

$t_n=5+\frac{9(4^{n-1}-1)}{4-1}$

$t_n=2+\frac{3}{16}{4}^n$ on simplification

$T_n=3+\sum t_n$ where number of terms are $(n-1)$

$T_n=3+2(n-1)+\frac{3}{16}\frac{4(4^{n-1}-1)}{4-1}$ ....$\left(7\right)$

$T_n=\frac{3}{4}+2n+\left(4^{n-2}\right)$ on simplification

$S_n=\sum T_n$ where number of terms are ${}^{\prime }{n}^{\prime }$

$S_n=\frac{3n}{4}+n(n+1)+\frac{4^n-1}{12}$