Prove by induction $\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$

Question

In this problem I have to prove the following equation using mathematical induction: $$\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$$

So far I've proved that $P_1$ is true, and written $P_k$ as $$15+36+63+...+3k(k+4)= \frac{(k)(k+1)(2k+13)}{2}$$

I then attempt to solve $P_{k+1}$. I start by using $P_k$ and substituting it into the equation to get $$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$

I then multiply $(3k+1)(k+5)$ by $\frac{2}{2}$ to get $$\frac{(k)(k+1)(2k+13)(3k+1)(2k+10)}{2}=\frac{(k+1)(k+2)(2k+14)}{2}$$

Here its obvious that the LHS isnt going to be equal to the RHS, seeing as the LHS would have $k^5$ amd the RHS would have $k^3$. Any ideas?

Answer 1

There are three silly mistakes here: First off, your expression on the RHS should be $\frac{(k+1)(k+2)(2k+15)}{2}$ since.

Secondly you somehow made the addition into multiplication here $$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$

And finally as Bernard said you should have a $3(k+1)=3k+3$ instead of $3k+1$ in your inductive step.

Try again starting from here $$\frac{(k)(k+1)(2k+13)}{2}+(3k+3)(k+5)=\frac{(k+1)(k+2)(2k+15)}{2}$$ and comment if you need more help.

Answer 2

Alternatively, prove a very useful identity: $$\sum_{k=0}^n\,\binom{k}{r}=\binom{n+1}{r+1}$$ for all integers $n,r\geq 0$. Then, note that $$3k(k+4)=6\binom{k}{2}+15\binom{k}{1}\,.$$ You can prove it by induction on $n$, or using some kind of combinatorial arguments. Then, we have $$\begin{align} \sum_{k=1}^n\,3k(k+4)&=\sum_{k=0}^n\,3k(k+4)=6\,\sum_{k=0}^n\,\binom{k}{2}+15\,\sum_{k=0}^n\,\binom{k}{1}=6\binom{n+1}{2+1}+15\binom{n+1}{1+1} \\&=6\binom{n+1}{3}+15\binom{n+1}{2}=(n+1)n(n-1)+\frac{15(n+1)n}{2}\\&=\frac{n(n+1)\big(2(n-1)+15\big)}{2}=\frac{n(n+1)(2n+13)}{2}\,.\end{align}$$

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Here is another proof, using the well known sum $$\sum_{k=0}^n\,k^2=\frac{n(n+1)(2n+1)}{6}\text{ for all integer }n\geq 0\,.$$ Note that $$\sum_{k=1}^n\,3k(k+4)=3\,\sum_{k=1}^n\,\big((k+2)^2-4\big)=3\,\left(\sum_{k=1}^n\,(k+2)^2-\sum_{k=1}^n\,4\right)\,.$$ That is, $$\sum_{k=1}^n\,3k(k+4)=3\,\left(\sum_{k=3}^{n+2}\,k^2-4n\right)=3\,\left(\sum_{k=0}^{n+2}\,k^2-5-4n\right)\,.$$ Now, plugging in $\sum_{k=0}^{n+2}\,k^2=\frac{(n+2)(n+3)(2n+5)}{6}$ yields $$\sum_{k=1}^n\,3k(k+4)=3\,\left(\frac{(n+2)(n+3)(2n+5)}{6}-5-4n\right)\,.$$ The rest is just some work with simple algebraic manipulations.

Answer 3

Only a comment!!! $\,\,$ (because your question is answered by others)

The simplest method and without induction comes from Discrete Mathematics.

$\displaystyle \sum\limits_{k=1}^n (ak^2 + bk) = \sum\limits_{k=1}^n (a (k+1)^{\underline{2}} + (b-a)k^{\underline{1}}) = a \frac{(n+2)^{\underline{3}}}{3} + (b-a)\frac{(n+1)^{\underline{2}}}{2} $

where $\,x^{\underline{n}}=x(x-1)...(x-n+1)$

Note: It looks similar to integration.