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4^(n+1) + 15n + 14 is divisible by 9 for every natural number n ge 0. |
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Answer» Solution :Let`P_n = 4^(n+1) + 15n + 14` When n = 1, `4^2 + 15 + 14 = 45` is divisible by 9. `therefore P_1` is true. Let `P_k` be true. i.e , `4^(k+1) + 15k + 14` is divisible by 9. Now `4^(k+1+1) + 15(k+1) + 14` = `4^(k+2) + 15k + 29` `4(k+1).4 + 60k + 56 - 45k - 27` = `4(4^(k+1) + 15k + 14) - 9(5k+3)` Which is divisible by 9. `therefore P_(k+1)` is true. `therefore P_n` is true for all VALUES of `n ge 0`. |
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