Question

Prove that the sum of n terms of the series

$11+103+1005+....$ is $\frac{10}{9}({10}^n-1)+n^2$

Answer 1

$S_n=11+103+1005+.......$ n terms

$S_n=(10+1)+(102+3)+(103+5)+.....+[10n+(2n-1\left)\right]$

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

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