Sum of the series `1 + 3 + 7 + 15 + 31 +… `to n terms is

Options

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

Answer

To find the sum of the series \( S_n = 1 + 3 + 7 + 15 + 31 + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the pattern in the series
The series is:
- \( 1 \) (which is \( 2^0 - 1 \))
- \( 3 \) (which is \( 2^2 - 1 \))
- \( 7 \) (which is \( 2^3 - 1 \))
- \( 15 \) (which is \( 2^4 - 1 \))
- \( 31 \) (which is \( 2^5 - 1 \))

We can observe that the \( n \)-th term can be expressed as:
\[ T_n = 2^n - 1 \]

### Step 2: Write the sum of the series in terms of \( T_n \)
Thus, the sum of the first \( n \) terms can be expressed as:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n = (2^1 - 1) + (2^2 - 1) + (2^3 - 1) + \ldots + (2^n - 1) \]

### Step 3: Simplify the sum
We can separate the sum:
\[ S_n = (2^1 + 2^2 + 2^3 + \ldots + 2^n) - n \]

### Step 4: Use the formula for the sum of a geometric series
The sum of a geometric series can be calculated using the formula:
\[ \text{Sum} = a \frac{r^n - 1}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

In our case:
- \( a = 2 \)
- \( r = 2 \)
- Number of terms = \( n \)

Thus, we have:
\[ 2^1 + 2^2 + 2^3 + \ldots + 2^n = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2 \]

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

### Final Answer
Thus, the sum of the series \( 1 + 3 + 7 + 15 + 31 + \ldots \) up to \( n \) terms is:
\[ S_n = 2^{n+1} - n - 2 \]

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