The greatest +ve integer k, for which `4^k+1` is factor of sum `1+2+2^2+....+2^(99)` is

Explanation

To solve the problem, we need to find the greatest positive integer \( k \) such that \( 4^k + 1 \) is a factor of the sum \( S = 1 + 2 + 2^2 + \ldots + 2^{99} \).

### Step 1: Calculate the sum of the series
The series \( S = 1 + 2 + 2^2 + \ldots + 2^{99} \) is a geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = 2 \)
- The number of terms \( n = 100 \)

The formula for the sum of a geometric series is given by:
\[ S = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values, we have:
\[ S = \frac{1(2^{100} - 1)}{2 - 1} = 2^{100} - 1 \]

### Step 2: Express \( S \) in terms of \( 4^k + 1 \)
We need to find \( k \) such that \( 4^k + 1 \) divides \( 2^{100} - 1 \). Notice that:
\[ 2^{100} - 1 = (2^{50})^2 - 1^2 = (2^{50} - 1)(2^{50} + 1) \]
We can further factor \( 2^{50} - 1 \):
\[ 2^{50} - 1 = (2^{25})^2 - 1^2 = (2^{25} - 1)(2^{25} + 1) \]

### Step 3: Relate \( 4^k + 1 \) to the factors of \( S \)
Now, we can express \( 2^{100} - 1 \) as:
\[ 2^{100} - 1 = (2^{25} - 1)(2^{25} + 1)(2^{50} + 1) \]
We know that \( 4^k + 1 = (2^{2k} + 1) \). We need to find the largest \( k \) such that \( 2^{2k} + 1 \) divides \( 2^{100} - 1 \).

### Step 4: Check divisibility
From the factorization, we observe:
- \( 2^{25} - 1 \) is divisible by \( 2^{2k} + 1 \) for \( k \leq 12 \) (since \( 2^{25} \equiv -1 \mod (2^{2k} + 1) \)).
- \( 2^{25} + 1 \) is not divisible by \( 2^{2k} + 1 \) for any \( k \).
- \( 2^{50} + 1 \) is divisible by \( 2^{2k} + 1 \) for \( k \leq 25 \).

### Step 5: Determine the maximum \( k \)
The maximum \( k \) such that \( 4^k + 1 \) divides \( 2^{100} - 1 \) is when \( 2^{2k} + 1 \) divides \( 2^{50} + 1 \). Since \( 2^{50} + 1 \) is the largest factor we can consider, we find:
\[ k = 25 \]

Thus, the greatest positive integer \( k \) for which \( 4^k + 1 \) is a factor of \( 2^{100} - 1 \) is:
\[ \boxed{25} \]