The sum `1+3+7+15+31+...to100` terms is `2^(100)-102 b` b. `2^(99)-101` c. `2^(101)-102` d. none of these

Options

A. `2^(100)-102`
B. `2^(99)-101`
C. `2^(101)-102`
D. none of these

Explanation

To solve the problem of finding the sum of the series \(1 + 3 + 7 + 15 + 31 + \ldots\) up to 100 terms, we can follow these steps:

### Step 1: Identify the Pattern
The given series is:
- \(1\)
- \(3\)
- \(7\)
- \(15\)
- \(31\)

We can observe that each term can be expressed in the form:
- \(1 = 2^1 - 1\)
- \(3 = 2^2 - 1\)
- \(7 = 2^3 - 1\)
- \(15 = 2^4 - 1\)
- \(31 = 2^5 - 1\)

Continuing this pattern, we can generalize the \(n\)-th term as:
\[ T_n = 2^n - 1 \]

### Step 2: Write the Sum of the Series
The sum of the first 100 terms can be expressed as:
\[ S = T_1 + T_2 + T_3 + \ldots + T_{100} \]
Substituting the expression for \(T_n\):
\[ S = (2^1 - 1) + (2^2 - 1) + (2^3 - 1) + \ldots + (2^{100} - 1) \]

### Step 3: Simplify the Sum
We can separate the terms:
\[ S = (2^1 + 2^2 + 2^3 + \ldots + 2^{100}) - (1 + 1 + 1 + \ldots + 1) \]
The second part is simply \(100\) (since there are 100 terms).

### Step 4: Calculate the Geometric Series
The first part is 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:
- \(a = 2\)
- \(r = 2\)
- \(n = 100\)

Thus, the sum becomes:
\[ 2 \frac{2^{100} - 1}{2 - 1} = 2(2^{100} - 1) = 2^{101} - 2 \]

### Step 5: Combine the Results
Now substituting back into our expression for \(S\):
\[ S = (2^{101} - 2) - 100 \]
\[ S = 2^{101} - 102 \]

### Conclusion
Thus, the sum of the series \(1 + 3 + 7 + 15 + 31 + \ldots\) up to 100 terms is:
\[ S = 2^{101} - 102 \]

The correct answer is option **C: \(2^{101} - 102\)**.