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.
 A: First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}(k+2)(k+4)=\frac{2\cdot1^3+21\cdot1^2+67\cdot1}{6}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}(k+2)(k+4)=\frac{2n^3+21n^2+67n}{6}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}(k+2)(k+4)=$
$\color\red{\sum\limits_{k=1}^{n}(k+2)(k+4)}+(n+1+2)(n+1+4)=$
$\color\red{\frac{2n^3+21n^2+67n}{6}}+(n+1+2)(n+1+4)=$
$\frac{2(n+1)^3+21(n+1)^2+67(n+1)}{6}$

Please note that the assumption is used only in the part marked red.
A: $(k+2)(k+4)=k^2+6k+8$
$$\sum^{n}_{k=1}{(k+2)(k+4)}=\sum^{n}_{k=1}{k^2}+6\sum^{n}_{k=1}{k}+\sum^{n}_{k=1}{8}=\frac{n(n+1)(2n+1)}{6}+6 \cdot \frac{n(n+1)}{2}+8n$$
A: You could also use that: 
$$(k+2)(k+4) = (k+3)^2 - 1$$
and the known sum:
$$ \sum^{n}_{k=1}{k^2} = \frac{n(n+1)(2n+1)}{6} $$
$$ \sum^{n}_{k=1}{(k+2)(k+4)} = \sum^{n}_{k=1}{((k+3)^2 - 1)} = \sum^{n}_{k=1}{(k+3)^2} - \sum^{n}_{k=1}{1} $$
$$ \sum^{n+3}_{k=4}{k^2} - n = \sum^{n+3}_{k=1}{k^2} - \sum^{3}_{k=1}{k^2} - n $$
$$ \frac{(n+3)(n+4)(2n+7)}{6} - 14 - n $$
The rest is simple.
A: 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.
A: As both members are cubic polynomials, it suffices to show equality for $4$ distinct values of $n$.
$$0=0=\frac{0+0+0}6,\\
0+3\cdot5=15=\frac{2+21+67}6,\\
0+3\cdot5+4\cdot6=39=\frac{2\cdot8+21\cdot4+67\cdot2}6,\\
0+3\cdot5+4\cdot6+5\cdot7=74=\frac{2\cdot27+21\cdot9+67\cdot3}6.\\
$$
QED.
