Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6
Best Answer
For any positive integer $n$, let $S(n)$ denote the statement $$ S(n) : 6\mid (n^3+11n). $$
Base step: For $n=1, S(1)$ gives $1^3+11(1) = 12 = 2\cdot 6$. Thus, $S(1)$ holds.
Inductive step: Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $6\ell = k^3+11k$. Then \begin{align} (k+1)^3 + 11(k+1) &= (k^3+3k^2+3k+1) + (11k+11)\tag{expand}\\[0.5em] &= \color{red}{k^3+11k}+3k^2+3k+12\tag{rearrange}\\[0.5em] &= \color{red}{6\ell} + 3k(k+1)+2\cdot 6.\tag{by ind. hyp.} \end{align} Since one of $k$ and $k+1$ is even, the term $3k(k+1)$ is divisible by $6$, and so the last expression above is divisible by $6$. This proves $S(k+1)$ and concludes the inductive step $S(k)\to S(k+1)$.
By mathematical induction, for each $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$