# Mathematical Induction: Divisibility by 7 Example

October 25, 2014 | 1 min read

Prove that $$3^{2n+1}+2^{n+2}$$ is divisible by $$7$$ for every nonnegative integer $$n$$.

Let $$P(n)$$ be: $$7\mid3^{2n+1}+2^{n+2}$$.

We wish to show that $$P(n)$$ is true $$\forall n \geq0$$.

We first verify the base $$n=0$$ case: $3^{2\times0+1}+2^{0+2}=3^1+2^2=7$ $1 \times 7 = 7$

We wish to show that $$P(k)\implies P(k+1) \, \forall k \geq 0$$.

We assume $$P(k)$$ to be true for some $$k \geq 0$$.

Thus $$7q=3^{2k+1}+2^{k+2}$$ for some $$q \in \mathbb{Z}$$.

Consider

\begin{align*} & 3^{2(k+1)+1}+2^{(k+1)+2} \\ &= 3^{(2k+1)+2}+2^{(k+2)+1} \\ &= 9(3^{2k+1})+2(2^{k+2}) \\ &= 9{(3^{2k+1}+2^{k+2})}-7(2^{k+2}) \end{align*}

We have shown $$3^{2(k+1)+1}+2^{(k+1)+2}=9(3^{2k+1}+2^{k+2})-7(2^{k+2})$$.

Because $$7\mid3^{2k+1}+2^{k+2}$$ as assumed, and $$7\mid7$$ trivially, it follows that $$7\mid3^{2(k+1)+1}+2^{(k+1)+2}$$.

Thus, $$P(k)\implies P(k+1)$$ as needed, so $$P(n)$$ is true $$\forall n \geq 0$$