Series in a series

let's go step by step:

• First step is to notice that each term forms a series in the order: 1,1+11,1+11+111,1+11+111+1111,..... and so on.

• Now, a $n^{th}$ term in the series can be defined as:

$a_{n}$ = 1+11+111+1111+.....n terms

= $\frac{1}{9}$ ( $9+99+999+9999....$ )

= $\frac{1}{9}$ ( $10-1+100-1+1000-1....$ )

= $\frac{1}{9}$ ( $10+100+1000+10000....-n$ )

• Now apply GP to obtain:

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

• Summate $a_{n}$ up to n terms:

$S_{n}$ = $\sum_{i=1}^n$ ( $\frac{1}{9}$ (10 $\frac{10^{n}-1}{9} - n$ ))

= $\frac{200(10^{n}-1)-180n-81(n)(n+1)}{1458}$

I leave it to you to compare the coefficients and sum them accordingly.