Question

$11,103,1005,.$Find the sum to $n$ terms of the series.

Answer 1

$11,103,1005\dots$

$(10+1),(1^2+3),({10}^3+5)\dots$

$[10+{10}^2+{10}^3\dots ]+[1+3+5+7\dots ]$

$G.P.a=10,r=10.$ $+$ $A.P.b=1,d=2.$

$S_n=S_1+S_2$

$S_n=\frac{a(r^n-1)}{r-1}+\frac{n}{2}[2b+(n-1\left)d\right]$

$S_n=\frac{10({10}^n-1)}{9}+\frac{n}{2}[2b+(n-1\left)2\right]$

$S_n=\frac{10}{9}({10}^n-1)+n[1+n-1]$

$S_n=\frac{10}{9}({10}^n-1)+n^2$