Sum of the series `1 + 2.2 + 3.2^(2) + 4.2^(3)+…+ 100.2^(99)` is

Options

A. `100.2^(100) + 1`
B. `99.2^(100) + 1`
C. `99.2^(99) - 1`
D. `100.2^(100) - 1`

Answer

To solve the series \( S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 100 \cdot 2^{99} \), we can use a technique involving multiplying the series by a constant and then manipulating the resulting equations.

### Step-by-step Solution:

1. **Define the Series**:
Let \( S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 100 \cdot 2^{99} \).

2. **Multiply the Series by 2**:
Multiply \( S \) by 2:
\[ 2S = 2 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 100 \cdot 2^{100} \]
This can be rewritten as:
\[ 2S = 2 + 2^2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 99 \cdot 2^{99} + 100 \cdot 2^{100} \]

3. **Align the Terms**:
Now, we can align the terms of \( S \) and \( 2S \):
\[ 2S = 2 + (2 + 3) \cdot 2^2 + (3 + 4) \cdot 2^3 + \ldots + 100 \cdot 2^{100} \]

4. **Subtract the Two Equations**:
Subtract \( S \) from \( 2S \):
\[ 2S - S = S = (2 + 2^2 + 2^2 + 2^3 + \ldots + 2^{99}) + 100 \cdot 2^{100} \]

5. **Simplify the Series**:
The series \( 1 + 2 + 3 + \ldots + 100 \) can be calculated using the formula for the sum of the first \( n \) natural numbers:
\[ \text{Sum} = \frac{n(n + 1)}{2} = \frac{100 \cdot 101}{2} = 5050 \]

6. **Combine the Results**:
Now, we have:
\[ S = 5050 \cdot 2^{99} + 100 \cdot 2^{100} \]

7. **Final Calculation**:
Rearranging gives:
\[ S = 5050 \cdot 2^{99} + 100 \cdot 2^{100} = 5050 \cdot 2^{99} + 100 \cdot 2 \cdot 2^{99} = (5050 + 200) \cdot 2^{99} = 5250 \cdot 2^{99} \]

### Final Answer:
Thus, the sum of the series is:
\[ S = 5250 \cdot 2^{99} \]