Use binomial theorem to show that $7^n + 2$ is divisible by $3$.
What I've done: $$\begin{aligned} (a+b)^n &= \sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r \\\\ 7^n+2 &=( 1+6)^n+2\\ &= \sum_{r=0}^{n}\dbinom{n}{r}1^{n-r}\times6^r+2 \\ &= \sum_{r=0}^{n}\dbinom{n}{r}6^r+2 \end{aligned}$$
But I'm not sure what to do from there. Would appreciate any help.
$$7^n + 2 = (6+1)^n + 2$$
Apply the binomial theorem to evaluate the first term on the RHS
$$\left(\sum_\limits{k=0}^n {n\choose k} 6^{k}1^{n-k}\right) + 2$$
In that summation every term where $k > 0$ we see that $3|6^{k}$ and hence $3\Bigg|\displaystyle{n\choose k} 6^{k}1^{n-k}$
When $k = 0$ we have $\displaystyle{n\choose 0} 6^{0}1^{n} = 1$
And $3|1+2$