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 \(11 + 103 + 1005 + \ldots\) is given by the formula \(\frac{10}{9}(10^n - 1) + n^2\), we will break down the series into two parts: a geometric progression (GP) and an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the Series**:
The series given is \(11 + 103 + 1005 + \ldots\). We can express the terms in a more manageable form:
- \(11 = 10^1 + 1\)
- \(103 = 10^2 + 3\)
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