Find the sum of n terms of the series 11 + 103 + 1005 + ...

Options

A. 10/9 (`10^n` – 1) – 1
B. 100/99(`10^n` – 1) + `n^2`
C. `10//9 (10^n - 1) + n^2`
D. none of these

Answer

To find the sum of the first n terms of the series 11 + 103 + 1005 + ..., we can break the terms down into two separate components: a geometric series and an arithmetic series. Let's solve it step by step.

### Step 1: Identify the Components of the Series
The series can be expressed as:
- The first term is 11, which can be rewritten as \(10 + 1\).
- The second term is 103, which can be rewritten as \(100 + 3\).
- The third term is 1005, which can be rewritten as \(1000 + 5\).

So, we can separate the series into two parts:
1. The series of the numbers \(10, 100, 1000, \ldots\)
2. The series of the numbers \(1, 3, 5, \ldots\)

### Step 2: Analyze the First Series
The first series \(10, 100, 1000, \ldots\) can be recognized as a geometric progression (GP) with:
- First term \(a = 10\)
- Common ratio \(r = 10\)

The sum of the first n terms of a geometric series is given by the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting the values:
\[ S_{GP} = 10 \frac{10^n - 1}{10 - 1} = 10 \frac{10^n - 1}{9} \]

### Step 3: Analyze the Second Series
The second series \(1, 3, 5, \ldots\) is an arithmetic progression (AP) where:
- First term \(a = 1\)
- Common difference \(d = 2\)

The sum of the first n terms of an arithmetic series is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values:
\[ S_{AP} = \frac{n}{2} \times (2 \cdot 1 + (n - 1) \cdot 2) = \frac{n}{2} \times (2 + 2n - 2) = \frac{n}{2} \times 2n = n^2 \]

### Step 4: Combine the Results
Now, we can find the total sum of the first n terms of the original series by adding the sums of the two components:
\[ S_n = S_{GP} + S_{AP} = 10 \frac{10^n - 1}{9} + n^2 \]

### Final Answer
Thus, the sum of the first n terms of the series is:
\[ S_n = \frac{10}{9} (10^n - 1) + n^2 \]

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