The largest value of the positive integer k for which `n^(k)+1` divides `1+n+n^(2)+ . . . .+n^(127)`, is

Options

A. 8
B. 16
C. 32
D. 64

Explanation

To find the largest value of the positive integer \( k \) for which \( n^k + 1 \) divides the sum \( 1 + n + n^2 + \ldots + n^{127} \), we can follow these steps:

### Step 1: Calculate the sum of the series
The series \( 1 + n + n^2 + \ldots + n^{127} \) is a geometric progression (GP) where:
- The first term \( a = 1 \)
- The common ratio \( r = n \)
- The number of terms \( n = 128 \) (from \( n^0 \) to \( n^{127} \))

The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values we have:
\[ S_{128} = \frac{1(n^{128} - 1)}{n - 1} = \frac{n^{128} - 1}{n - 1} \]

### Step 2: Factor the sum
Next, we can factor \( n^{128} - 1 \) using the difference of squares:
\[ n^{128} - 1 = (n^{64} - 1)(n^{64} + 1) \]
Thus, we can rewrite the sum as:
\[ S_{128} = \frac{(n^{64} - 1)(n^{64} + 1)}{n - 1} \]

### Step 3: Identify divisibility by \( n^k + 1 \)
We need to find the largest \( k \) such that \( n^k + 1 \) divides \( S_{128} \). From our factorization, we see that:
\[ S_{128} = \frac{(n^{64} - 1)(n^{64} + 1)}{n - 1} \]
The term \( n^{64} + 1 \) is present in the factorization.

### Step 4: Compare \( n^{k} + 1 \) with \( n^{64} + 1 \)
For \( n^k + 1 \) to divide \( S_{128} \), we can set:
\[ n^k + 1 = n^{64} + 1 \]
This implies that \( k = 64 \).

### Conclusion
Thus, the largest value of the positive integer \( k \) for which \( n^k + 1 \) divides the sum \( 1 + n + n^2 + \ldots + n^{127} \) is:
\[ \boxed{64} \]