<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] Another method (maybe overkill :D): The sum $S(n)=\sum_{k=1}^nP(k)$ in which $P(k)$ is a polynomial of $l^{th}$ degree can be expressed by a polynomial in $n$ of degree $l+1$. For your problem $P(k)=(k+2)(k+4)$ and has degree 2. Hence the sum can be expressed as a polynomial $G(n)=a_3n^3+a_2n^2+a_1n+a_0$. Now calculate four terms in your sum and solve the system $G(i)=a_3i^3+a_2i^2+a_1i+a_0=S(i)$. This method is more powerful, as it allows you to derive arbitrary sums over polynomial expresssions.
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