AP EAMCET · Maths · Basic of Mathematics
The least positive integer greater than 1 that divides \(49^n+16 n-1\) for all positive integers \(n\) is
- A \(64\)
- B \(49\)
- C \(7\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(P(n) = 49^n+16n-1\) \(P(n) = (1+48)^n+16n-1\) \(P(n) = 1 + n \cdot 48 + \binom{n}{2} 48^2 + \dots + 48^n + 16n-1\) \(P(n) = 64n + \binom{n}{2} 48^2 + \dots + 48^n\) Since \(48^2 = (3 \cdot 16)^2 = 9 \cdot 256 = 36 \cdot 64\), \(48^k\) is divisible by \(64\) for \(k \ge 2\).…
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