9231 P11 - Nov 2022 - Q02
4247
Prove by mathematical induction that, for all positive integers n, \(7^{2n} + 97^n - 50\) is divisible by 48. [6]
Solution
Checked by expert
1. Base Case: For \(n = 1\), \(7^{2 \times 1} + 97^1 - 50 = 49 + 97 - 50 = 96\), which is divisible by 48.
2. Inductive Hypothesis: Assume \(7^{2k} + 97^k - 50\) is divisible by 48 for some positive integer \(k\).
3. Inductive Step: Consider \(7^{2(k+1)} + 97^{k+1} - 50\).
4. \(7^{2(k+1)} + 97^{k+1} - 50 = 7^{2k+2} + 97^{k+1} - 50 = (48+1)7^{2k} + (96+1)97^k - 50\).
5. This expression is divisible by 48 because \(48 \times 7^{2k} + 96 \times 97^k\) is divisible by 48.
6. Therefore, by induction, \(7^{2n} + 97^n - 50\) is divisible by 48 for all positive integers \(n\).
