The sum of the series `(1+2) +(1+2+2^(2))+ (1+2+2^(2)+2^(3)) +….` upto n terms is

Options

A. ` 2^(n+2)-n-4`
B. ` 2 (2^(n)-1)-n`
C. ` 2^(n+1)-n`
D. ` 2^(n+1)-1`

Explanation

To find the sum of the series \( (1 + 2) + (1 + 2 + 2^2) + (1 + 2 + 2^2 + 2^3) + \ldots \) up to \( n \) terms, we can break it down step by step.

### Step 1: Identify the pattern in the series
The series can be expressed as:
- The first term is \( 1 + 2 \)
- The second term is \( 1 + 2 + 2^2 \)
- The third term is \( 1 + 2 + 2^2 + 2^3 \)
- And so on...

The \( k \)-th term of the series can be represented as:
\[ T_k = 1 + 2 + 2^2 + \ldots + 2^{k-1} \]

### Step 2: Use the formula for the sum of a geometric series
The sum of the first \( k \) terms of a geometric series can be calculated using the formula:
\[ S_k = a \frac{r^k - 1}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( k \) is the number of terms.

For our series:
- \( a = 1 \)
- \( r = 2 \)
- The number of terms in \( T_k \) is \( k \).

Thus, we have:
\[ T_k = 1 \cdot \frac{2^k - 1}{2 - 1} = 2^k - 1 \]

### Step 3: Write the total sum of the series up to \( n \) terms
Now, we need to find the sum of the first \( n \) terms:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
Substituting the expression for \( T_k \):
\[ S_n = (2^1 - 1) + (2^2 - 1) + (2^3 - 1) + \ldots + (2^n - 1) \]

### Step 4: Simplify the sum
This can be rewritten as:
\[ S_n = (2^1 + 2^2 + 2^3 + \ldots + 2^n) - n \]

The sum \( 2^1 + 2^2 + 2^3 + \ldots + 2^n \) is another geometric series with:
- First term \( a = 2 \)
- Common ratio \( r = 2 \)
- Number of terms \( n \)

Using the geometric series sum formula:
\[ S = a \frac{r^n - 1}{r - 1} = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2 \]

### Step 5: Final expression for \( S_n \)
Now substituting back into our equation for \( S_n \):
\[ S_n = (2^{n+1} - 2) - n = 2^{n+1} - n - 2 \]

Thus, the sum of the series up to \( n \) terms is:
\[ \boxed{2^{n+1} - n - 2} \]