Find the sum to n terms of
`3 * 2 + 5*2^2 + 7*2^3 + ….`

Options

A. `2^(n+2) - 2^(n+1)-2`
B. `n* 2^(n+2) - 2^(n+1) + 2`
C. `n*2^(n+2) + 2^(n-2)`
D. none of these

Answer

To find the sum to n terms of the series \(3 \cdot 2 + 5 \cdot 2^2 + 7 \cdot 2^3 + \ldots\), we can follow these steps:

### Step 1: Identify the pattern in the series
The series can be expressed as:
\[ S_n = 3 \cdot 2^1 + 5 \cdot 2^2 + 7 \cdot 2^3 + \ldots \]
We can see that the coefficients of \(2^n\) are odd numbers starting from 3. The general term can be represented as:
\[ a_n = (2n + 1) \cdot 2^n \]
where \(n\) starts from 1.

### Step 2: Write the sum of the first n terms
The sum of the first n terms can be written as:
\[ S_n = \sum_{k=1}^{n} (2k + 1) \cdot 2^k \]

### Step 3: Split the sum into two parts
We can separate the sum into two parts:
\[ S_n = \sum_{k=1}^{n} 2k \cdot 2^k + \sum_{k=1}^{n} 2^k \]
This simplifies to:
\[ S_n = 2 \sum_{k=1}^{n} k \cdot 2^k + \sum_{k=1}^{n} 2^k \]

### Step 4: Calculate the second sum
The second sum \(\sum_{k=1}^{n} 2^k\) is a geometric series:
\[ \sum_{k=1}^{n} 2^k = 2(1 + 2 + 2^2 + \ldots + 2^{n-1}) = 2 \left(\frac{2^n - 1}{2 - 1}\right) = 2(2^n - 1) = 2^{n+1} - 2 \]

### Step 5: Calculate the first sum
To calculate \(\sum_{k=1}^{n} k \cdot 2^k\), we can use the formula:
\[ \sum_{k=1}^{n} kx^k = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) \]
where \(x = 2\). The sum \(\sum_{k=0}^{n} x^k = \frac{x^{n+1} - 1}{x - 1}\).

Differentiating:
\[ \sum_{k=1}^{n} kx^k = x \frac{d}{dx} \left( \frac{x^{n+1} - 1}{x - 1} \right) \]

Calculating this gives:
\[ \sum_{k=1}^{n} k \cdot 2^k = 2 \cdot \frac{(n+1)2^{n+1} - 2^{n+1} + 1}{1} = (n - 1)2^{n+1} + 2 \]

### Step 6: Combine the results
Now, substituting back into our expression for \(S_n\):
\[ S_n = 2 \left((n - 1)2^{n+1} + 2\right) + (2^{n+1} - 2) \]
This simplifies to:
\[ S_n = 2(n - 1)2^{n+1} + 4 + 2^{n+1} - 2 = (2n + 1)2^{n+1} + 2 \]

### Final Result
Thus, the sum to n terms of the series is:
\[ S_n = (2n + 1)2^{n+1} + 2 \]