Step 1: Understanding the Question:
We need to simplify the given expression using the formula for the sum of a geometric progression (GP) and then analyze its factors.
Step 2: Applying the Method:
The sum of a geometric series is \( S_n = a\frac{r^n - 1}{r - 1} \).
The term \( x^n - y^n \) is divisible by \( x - y \).
The term \( x^n - y^n \) contains \( x^{n/2} + y^{n/2} \) as a factor if \( n \) is even.
Step 3: Computation Steps:
Let the given expression be \( E \).
\[ E = 48(1 + 49 + 49^2 + \dots + 49^{125}) \]
The term in the bracket is a GP with first term \( a=1 \), common ratio \( r=49 \), and number of terms \( n=126 \) (powers from 0 to 125).
\[ \text{Sum} = \frac{49^{126} - 1}{49 - 1} = \frac{49^{126} - 1}{48} \]
Substitute this back into the expression for \( E \):
\[ E = 48 \left( \frac{49^{126} - 1}{48} \right) = 49^{126} - 1 \]
We are looking for \( k \) such that \( 49^k + 1 \) is a factor of \( 49^{126} - 1 \).
Recall the identity \( a^2 - b^2 = (a - b)(a + b) \).
Let \( a = 49^{63} \). Then:
\[ 49^{126} - 1 = (49^{63})^2 - 1^2 = (49^{63} - 1)(49^{63} + 1) \]
From this, it is clear that \( 49^{63} + 1 \) is a factor of \( 49^{126} - 1 \).
Therefore, \( k = 63 \) works.
We must check if a larger \( k \) is possible. The factors of the form \( 49^k + 1 \) for \( 49^{126}-1 \) arise from the factorization \( x^{2m} - 1 = (x^m-1)(x^m+1) \). The term \( x^m+1 \) is a factor. Here \( 2m = 126 \implies m=63 \).
Further splitting \( 49^{63}-1 \) or \( 49^{63}+1 \) would yield factors with lower powers or odd powers not in the form \( 49^k+1 \) with \( k>63 \). Thus, 63 is the greatest such integer.
Step 4: Required Answer:
The greatest positive integer \( k \) is 63.