The sum of first 50 term of the series `1+3/2+7/4+15/8+31/16+`….. Is `(p+1/(2^(q)))`, then value of `(p+q)` is …….

Explanation

To find the sum of the first 50 terms of the series \(1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \ldots\), we can analyze the series and derive a formula for the \(n^{th}\) term.

### Step 1: Identify the pattern in the series
The series can be expressed as:
\[ \frac{1}{2^0} + \frac{3}{2^1} + \frac{7}{2^2} + \frac{15}{2^3} + \frac{31}{2^4} + \ldots \]

We can observe that the numerators follow a pattern:
- \(1 = 2^1 - 1\)
- \(3 = 2^2 - 1\)
- \(7 = 2^3 - 1\)
- \(15 = 2^4 - 1\)
- \(31 = 2^5 - 1\)

Thus, the \(n^{th}\) term can be expressed as:
\[ a_n = \frac{2^{n+1} - 1}{2^n} \]

### Step 2: Rewrite the \(n^{th}\) term
We can rewrite \(a_n\) as:
\[ a_n = 2 - \frac{1}{2^n} \]

### Step 3: Sum the first 50 terms
Now, we need to find the sum of the first 50 terms:
\[ S_{50} = \sum_{n=0}^{49} a_n = \sum_{n=0}^{49} \left(2 - \frac{1}{2^n}\right) \]
This can be split into two separate sums:
\[ S_{50} = \sum_{n=0}^{49} 2 - \sum_{n=0}^{49} \frac{1}{2^n} \]

### Step 4: Calculate the first sum
The first sum is:
\[ \sum_{n=0}^{49} 2 = 2 \times 50 = 100 \]

### Step 5: Calculate the second sum
The second sum is a geometric series:
\[ \sum_{n=0}^{49} \frac{1}{2^n} = \frac{1 - \left(\frac{1}{2}\right)^{50}}{1 - \frac{1}{2}} = 2 \left(1 - \frac{1}{2^{50}}\right) \]

### Step 6: Combine the results
Now, substituting back into the sum:
\[ S_{50} = 100 - 2 \left(1 - \frac{1}{2^{50}}\right) = 100 - 2 + \frac{2}{2^{50}} = 98 + \frac{1}{2^{49}} \]

### Step 7: Express in the required form
We have:
\[ S_{50} = 98 + \frac{1}{2^{49}} \]
This matches the form \(p + \frac{1}{2^q}\) where \(p = 98\) and \(q = 49\).

### Step 8: Calculate \(p + q\)
Now, we find:
\[ p + q = 98 + 49 = 147 \]

Thus, the final answer is:
\[ \boxed{147} \]