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3^(2(n-1)) + 7 is divisible by 8 for eve...

`3^(2(n-1)) + 7` is divisible by 8 for every natural `n ge 2`.

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Let `P_n = 3^(2(n-1)) + 7`
When `n =2. 3^2 + 7 = 16` is divisible by 8.
`therefore P_2` is true.
`P_k` be true. i.e, `3^(2(k-1)) + 7` is divisible y 8. Let `3^(2k-2) + 7` = 8m. M in Z . Now `3^(2(k+1-1) +7 = 3^2k + 7`
`3^(2k-2).3^2 + 63-56`
= `9(3(2k-2) + 7)-56`
= `9 xx 8m - 56` = 8(9m-7) Which is divisible by 8.
`therefore P_(k+1)` is true.
`therefore P_n` is true for all values on `n le 2`
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