<s>[INST] Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this: 

Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the  $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$  $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$  $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
 [/INST] $(k+2)(k+4)=k^2+6k+8$
$$\sum^{n}_{k=1}{(k+2)(k+4)}=\sum^{n}_{k=1}{k^2}+6\sum^{n}_{k=1}{k}+\sum^{n}_{k=1}{8}=\frac{n(n+1)(2n+1)}{6}+6 \cdot \frac{n(n+1)}{2}+8n$$
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