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4^(n+1) + 15n + 14 is divisible by 9 for...

`4^(n+1) + 15n + 14` is divisible by 9 for every natural number `n ge 0`.

Text Solution

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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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