For all natural numbers `5^(2k - 1) + 1` is always divisible by 6 can be proved using mathematical induction.

First take k = 1, we have `5^(2 - 1) + 1 = 5 + 1 = 6` which is divisible by 6.

Assume the assertion is true for a number k; we have to show if that is true, it also holds for k + 1.

The assumption gives: `5^(2k - 1) + 1 ` = 6*n

Now, `5^(2(k + 1) - 1) + 1`

=> `5^(2k + 2 - 1) + 1`

=> `5^(2k - 1 + 2) + 1`

=> `5^(2k - 1)*25 + 1`

=> `5^(2k - 1)(24 + 1) + 1`

=> `5^(2k - 1)*24 + 5^(2k - 1) + 1`

but `5^(2k - 1) + 1 = 6n`

=> `6n + 5^(2k - 1)*24`

as 24 is divisible by 6, both the terms are divisible by 6.

**This proves by mathematical induction that for all natural numbers k, **`5^(2k - 1) + 1` **are divisible by 6.**

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