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

Explanation

To prove that the sum of the series \( S_n = 11 + 103 + 1005 + \ldots \) can be expressed as \( S_n = \frac{10}{9}(10^n - 1) + n^2 \), we will break down the series into two parts: a geometric progression (GP) and an arithmetic progression (AP).

### Step 1: Identify the Series Components
The given series can be rewritten as:
\[ S_n = (11 + 103 + 1005 + \ldots) \]
This series can be split into two parts:
1. The terms that are powers of 10: \( 10 + 10^2 + 10^3 + \ldots \)
2. The constant terms: \( 1 + 3 + 5 + \ldots \)
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