Sum to n-terms of the series `(1/2)^2 + (3/4)^2 + (7/8)^2 + (15/16)^2` +…. Is given by :

Options

A. `n-1/3 (4^(-n) -5) + 2^(1-n)`
B. `n-1/3 (4^(-n) + 5) + 2^(-n)`
C. `n-1/3 (4^(-n) + 5) +2^(1-n)`
D. None of these

Answer

To find the sum to n-terms of the series

\[ \left(\frac{1}{2}\right)^2 + \left(\frac{3}{4}\right)^2 + \left(\frac{7}{8}\right)^2 + \left(\frac{15}{16}\right)^2 + \ldots \]

we can start by identifying a pattern in the series.

### Step 1: Identify the general term of the series

The terms can be rewritten as:

\[ \left(1 - \frac{1}{2}\right)^2, \left(1 - \frac{1}{4}\right)^2, \left(1 - \frac{1}{8}\right)^2, \left(1 - \frac{1}{16}\right)^2 \]

This suggests a pattern where the \(n\)-th term can be expressed as:

\[ \left(1 - \frac{1}{2^n}\right)^2 \]

### Step 2: Expand the general term

Now, we expand the general term:

\[ \left(1 - \frac{1}{2^n}\right)^2 = 1 - 2 \cdot \frac{1}{2^n} + \left(\frac{1}{2^n}\right)^2 \]

### Step 3: Write the sum of n terms

The sum of the first \(n\) terms of the series can be expressed as:

\[ S_n = \sum_{k=1}^{n} \left(1 - 2 \cdot \frac{1}{2^k} + \frac{1}{4^k}\right) \]

### Step 4: Separate the sum

This can be separated into three parts:

\[ S_n = \sum_{k=1}^{n} 1 - 2 \sum_{k=1}^{n} \frac{1}{2^k} + \sum_{k=1}^{n} \frac{1}{4^k} \]

### Step 5: Calculate each sum

1. The first sum is simply \(n\):

\[ \sum_{k=1}^{n} 1 = n \]

2. The second sum is a geometric series:

\[ \sum_{k=1}^{n} \frac{1}{2^k} = \frac{\frac{1}{2}(1 - \left(\frac{1}{2}\right)^n)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} \]

3. The third sum is also a geometric series:

\[ \sum_{k=1}^{n} \frac{1}{4^k} = \frac{\frac{1}{4}(1 - \left(\frac{1}{4}\right)^n)}{1 - \frac{1}{4}} = \frac{1}{3}\left(1 - \frac{1}{4^n}\right) \]

### Step 6: Combine the results

Now substituting back into \(S_n\):

\[ S_n = n - 2\left(1 - \frac{1}{2^n}\right) + \frac{1}{3}\left(1 - \frac{1}{4^n}\right) \]

### Step 7: Simplify the expression

Combining all the terms:

\[ S_n = n - 2 + \frac{2}{2^n} + \frac{1}{3} - \frac{1}{3 \cdot 4^n} \]

This can be simplified to:

\[ S_n = n - \frac{5}{3} + \frac{2}{2^n} - \frac{1}{3 \cdot 4^n} \]

### Final Result

Thus, the sum to n-terms of the series is:

\[ S_n = n - \frac{5}{3} + \frac{2}{2^n} - \frac{1}{3 \cdot 4^n} \]