Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$

Question

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:

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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.

Answer 1

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 $n+1$:

$\sum\limits_{k=1}^{n+1}(k+2)(k+4)=$

$\color\red{\sum\limits_{k=1}^{n}(k+2)(k+4)}+(n+1+2)(n+1+4)=$

$\color\red{\frac{2n^3+21n^2+67n}{6}}+(n+1+2)(n+1+4)=$

$\frac{2(n+1)^3+21(n+1)^2+67(n+1)}{6}$

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Please note that the assumption is used only in the part marked red.

Answer 2

$(k+2)(k+4)=k^2+6k+8$

$$\sum^{n}_{k=1}{(k+2)(k+4)}=\sum^{n}_{k=1}{k^2}+6\sum^{n}_{k=1}{k}+\sum^{n}_{k=1}{8}=\frac{n(n+1)(2n+1)}{6}+6 \cdot \frac{n(n+1)}{2}+8n$$

Answer 3

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.

Answer 4

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.

Answer 5

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.