Prove that the sum to `n` terms of the series `11+103+1005+ i s(10//9)(10^n-1)+n^2dot`

Explanation

Let `S_(n)` denote the sum to n terms of the given series.
Then, `S_(n)=11+103+1005+…` to n terms
`=(10+1)+(10^(2)+3)+(10^(3)+5)+…+{10^(n)+(2n-1)}`
`=underset(G.P.)ubrace((10+10^(2)+...+10^(n)))+underset(A.P.)ubrace({1+3+5+..+(2n-1)})`
`=(10(10^(n)-1))/((10-1))+n/2{1+(2n-1)}`
`=10/9(10^(n)-1)+n^(2)`