Show that for every positive integer n, 7 divides 3(to the power 2n+1) + 2(to the power n+2)

Working mod 7,
3^(2n+1) + 2^(n+2)
= 3 * 9^n + 4 * 2^n
= 3 * 2^n + 4 * 2^n
= (3 + 4) * 2^n
= 0.

Hence, 7 | (3^(2n+1) + 2^(n+2)).
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Alternately, use Mathematical Induction.

Base Case (n = 1):
This is true for n = 1, because 3^(2+1) + 2^(1+2) = 27 + 8 = 35, which is divisible by 7.

Inductive Step:
Assuming that this is true for n = k:

3^(2(k+1) + 1) + 2^((k+1) + 2)
= 3^(2k+3) + 2^(k+3)
= 3^2 * 3^(2k+1) + 2 * 2^(k+2)
= 9 * 3^(2k+1) + 2 * 2^(k+2)
= 7 * 3^(2k+1) + 2 [3^(2k+1) + 2^(k+2)].

Clearly, 7 | 7 * 3^(2k+1).
Moreover, 7 | [3^(2k+1) + 2^(k+2)], by the inductive hypothesis.
Hence, 7 | [3^(2k+3) + 2^(k+3)], establishing the inductive step.

I hope this helps!