Sum of n terms of series `12+16+24+40+....` (A) `2(2^n -1)+8n` (B) `2(2^n-1)+6n` (C) `3(2^n-1)+8n` (D) `4(2^n-1)+8n`

Explanation

To find the sum of the first \( n \) terms of the series \( 12 + 16 + 24 + 40 + \ldots \), we can follow these steps:

### Step 1: Identify the pattern in the series
The given series is \( 12, 16, 24, 40, \ldots \). Let's find the differences between consecutive terms:
- \( 16 - 12 = 4 \)
- \( 24 - 16 = 8 \)
- \( 40 - 24 = 16 \)

The differences are \( 4, 8, 16 \), which are \( 4 \times 1, 4 \times 2, 4 \times 4 \). This indicates that the differences are increasing in a pattern of powers of 2 multiplied by 4.

### Step 2: Express the \( n \)-th term
From the pattern observed, we can express the \( n \)-th term \( T_n \) as follows:
- The first term \( T_1 = 12 \)
- The second term \( T_2 = 16 = 12 + 4 \)
- The third term \( T_3 = 24 = 16 + 8 \)
- The fourth term \( T_4 = 40 = 24 + 16 \)

We can see that:
\[ T_n = 12 + 4(1 + 2 + 4 + \ldots + 2^{n-2}) \]
where \( 1 + 2 + 4 + \ldots + 2^{n-2} \) is a geometric series.

### Step 3: Sum of the geometric series
The sum of the first \( m \) terms of a geometric series can be calculated using the formula:
\[ S_m = a \frac{r^m - 1}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( m \) is the number of terms.

In our case:
- \( a = 1 \)
- \( r = 2 \)
- \( m = n-1 \)

Thus, the sum becomes:
\[ S_{n-1} = 1 \cdot \frac{2^{n-1} - 1}{2 - 1} = 2^{n-1} - 1 \]

### Step 4: Substitute back to find \( T_n \)
Now substituting back into the expression for \( T_n \):
\[ T_n = 12 + 4(2^{n-1} - 1) = 12 + 4 \cdot 2^{n-1} - 4 = 8 + 4 \cdot 2^{n-1} \]

### Step 5: Find the sum of the first \( n \) terms \( S_n \)
The sum of the first \( n \) terms \( S_n \) is given by:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
Using the expression for \( T_n \):
\[ S_n = \sum_{k=1}^{n} (8 + 4 \cdot 2^{k-1}) = \sum_{k=1}^{n} 8 + \sum_{k=1}^{n} 4 \cdot 2^{k-1} \]
The first sum is simply \( 8n \), and the second sum is:
\[ 4 \sum_{k=0}^{n-1} 2^k = 4 \cdot \frac{2^n - 1}{2 - 1} = 4(2^n - 1) \]

### Final Step: Combine the results
Thus, we have:
\[ S_n = 8n + 4(2^n - 1) = 4(2^n - 1) + 8n \]

### Conclusion
The sum of the first \( n \) terms of the series is:
\[ S_n = 4(2^n - 1) + 8n \]

### Answer
The correct option is \( \text{(D) } 4(2^n - 1) + 8n \).