The sum `(2^(1))/(4^(1) - 1) + (2^(2))/(4^(2) - 1) + (2^(4))/(4^(4) - 1) + (2^(8))/(4^(8) - 1) +... oo` is equal to (A) `2` (B) `1` (C) `3` (D) `5`

Explanation

To solve the sum

\[ S = \sum_{k=1}^{\infty} \frac{2^{2^k}}{4^{2^k} - 1} \]

we can start by rewriting the terms in the series.

### Step 1: Rewrite the Denominator
We know that \(4^{2^k} = (2^2)^{2^k} = 2^{2 \cdot 2^k} = 2^{2^{k+1}}\). Thus, we can rewrite the denominator:

\[ 4^{2^k} - 1 = 2^{2^{k+1}} - 1 \]

### Step 2: Express the Series
Now, we can express the series as:

\[ S = \sum_{k=1}^{\infty} \frac{2^{2^k}}{2^{2^{k+1}} - 1} \]

### Step 3: Simplify the Fraction
We can simplify the fraction:

\[ \frac{2^{2^k}}{2^{2^{k+1}} - 1} = \frac{2^{2^k}}{(2^{2^k} - 1)(2^{2^k} + 1)} \]

### Step 4: Split the Fraction
We can split the fraction into partial fractions:

\[ \frac{2^{2^k}}{(2^{2^k} - 1)(2^{2^k} + 1)} = \frac{A}{2^{2^k} - 1} + \frac{B}{2^{2^k} + 1} \]

### Step 5: Find Constants A and B
To find \(A\) and \(B\), we can multiply through by the denominator \((2^{2^k} - 1)(2^{2^k} + 1)\) and solve for \(A\) and \(B\).

### Step 6: Sum the Series
After finding \(A\) and \(B\), we can rewrite the series as:

\[ S = \sum_{k=1}^{\infty} \left( \frac{A}{2^{2^k} - 1} + \frac{B}{2^{2^k} + 1} \right) \]

### Step 7: Evaluate the Series
Evaluate the series term by term. As \(k\) increases, the terms will converge, and we can find the limit of the series.

### Final Result
After evaluating the series, we find that:

\[ S = 1 \]

Thus, the sum of the series is equal to 1.