<s>[INST] Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$ 
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \times 2 = 7 \times 3^{2k+1} + 2 \times 3^{2k+1} + 2^{k-1} \times 2.$$
Where do I go from here?
 [/INST] The correct expression is obviously $$3^{2n+1}+ 2^{n-1}$$.
$n=1: 3^3 +2^0 = 27 +1 =28$
$n=2: 3^5 +2^1 = 245 = (35)(7)$
$n=3: 3^7 +2^2 = 2191 = (313)(7)$
So
 3^[2(n+1)+1] +2^(n+1-1) = [3^(2n+1)](9) + 2^n

                    = [7k-2^(n-1)](9) +2^n

                    = 63k -2^(n-1)(9-2)

                    = (7)[9k - 2^(n-1)]

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