Find the greatest value of k for which `49^(k)+1` is a factor of `1+49+49^(2)….(49)^(125)`
A
63
B
65
C
2
D
5
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to find the greatest value of \( k \) for which \( 49^k + 1 \) is a factor of the sum \( 1 + 49 + 49^2 + \ldots + 49^{125} \).
### Step-by-Step Solution:
1. **Identify the Series**: The expression \( 1 + 49 + 49^2 + \ldots + 49^{125} \) is a geometric series.
2. **Use the Geometric Series Formula**: The sum \( S_n \) of the first \( n \) terms of a geometric series can be calculated using the formula:
\[
S_n = \frac{a(r^n - 1)}{r - 1}
\]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.
Here, \( a = 1 \), \( r = 49 \), and \( n = 126 \) (since we have terms from \( 0 \) to \( 125 \)).
3. **Calculate the Sum**:
\[
S = \frac{1(49^{126} - 1)}{49 - 1} = \frac{49^{126} - 1}{48}
\]
4. **Factor the Numerator**: We can factor \( 49^{126} - 1 \) using the difference of squares:
\[
49^{126} - 1 = (49^{63} - 1)(49^{63} + 1)
\]
5. **Substituting Back**: Now we can rewrite the sum:
\[
S = \frac{(49^{63} - 1)(49^{63} + 1)}{48}
\]
6. **Finding the Factor**: We need to determine when \( 49^k + 1 \) is a factor of \( S \). The term \( 49^{63} + 1 \) is crucial here.
For \( 49^k + 1 \) to be a factor of \( 49^{63} + 1 \), we need \( k \) to be a divisor of \( 63 \).
7. **Divisors of 63**: The divisors of \( 63 \) are \( 1, 3, 7, 9, 21, 63 \).
8. **Greatest Value of \( k \)**: The greatest divisor of \( 63 \) is \( 63 \) itself.
### Conclusion:
Thus, the greatest value of \( k \) for which \( 49^k + 1 \) is a factor of \( 1 + 49 + 49^2 + \ldots + 49^{125} \) is:
\[
\boxed{63}
\]
