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This is a very interesting problem that I came across in an old textbook of mine. So I know its got something to do with mathematical induction, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Prove that for $n ≥ 1,$

$1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$

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  • $\begingroup$ Does the proof have to be by induction? $\endgroup$ – Sri-Amirthan Theivendran May 28 '16 at 21:20
  • $\begingroup$ By no means, no, however that is the way I approached it. $\endgroup$ – anonymous May 28 '16 at 21:22
  • $\begingroup$ You have formulas available for the sum of integers and the sum of squares of integers. Your sum is equivalent to $ \ \sum_{k=1}^n \ k^2 \ \ + \ \ 4 \ \sum_{k=1}^n \ k \ $ . $\endgroup$ – colormegone May 28 '16 at 21:34
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Base Case: For $n = 1$, we have: $$ 1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13) $$ which works.

Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$.

It remains to show that the claim holds for $n' = n$. Indeed, observe that: \begin{align*} &1 \times 5 +\cdots + n(n + 4) & \\ &= [1 \times 5 + \cdots + (n - 1)(n + 3)] + n(n + 4) &\text{since } n \geq 2 \\ &= \frac{1}{6}(n - 1)(n)(2n + 11) + n(n + 4) &\text{by the inductive hypothesis} \\ &= \frac{1}{6}n [(n - 1)(2n + 11) + 6(n + 4)] \\ &= \frac{1}{6}n [(2n^2 + 9n - 11) + (6n + 24)] \\ &= \frac{1}{6}n [2n^2 + 15n + 13] \\ &= \frac{1}{6}n (n + 1)(2n + 13) \\ \end{align*} as desired. $~~\blacksquare$

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A proof without induction: $$ \sum_{k=1}^{n}k(k+4)=\sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k= \frac{n(n+1)(2n+1)}{6}+4\cdot\frac{n(n+1)}{2} =\frac{n(n+1)}{6}(2n+13) $$

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$$1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = $$ $$=\sum_{i=1}^{n}i^2+\sum_{i=1}^{n}4i=\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}$$

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With induction, you want to prove a base case and then show that if it works for step $n$ then it also works for step $n+1$. If both of those things are true then you're done!

Base case, $n=1$:

$$ 1(1+4) = \frac{1}{6}1(1+1)(2(1)+13) $$ $$ 5 = 5 $$

Inductive step:

We want to show $$ 1 \times 5 + ... + (n+1)((n+1)+4) = \frac{1}{6}(n+1)((n+1)+1)(2(n+1)+13) $$ so we have $$ 1 \times 5 + ... + (n+1)((n+1)+4) = \frac{1}{6}(n+1)((n+1)+1)(2(n+1)+13) $$ $$ 1 \times 5 + ... + (n+1)((n+1)+4) = \frac{1}{6}(n+1)(n+2)(2n+15) $$ $$ \sum_{k=1}^{n+1} k(k+4) = \frac{1}{6}(n+1)(n+2)(2n+15) $$ Can you show that the series is equal to the right side?

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Induction is usually straight forward:

1) n = 1

Does $1*5 = \frac 16*1*(1+1)*(2*1 + 13)$?

Yep $5 = \frac 16 2*15 = 5$. So that's the initial step.

2) Suppose true for $n = k$

Does $1*5 + ..... + k(k+4) + (k+1)(k+5) = $

$\frac 16*k(k+1)(2k + 13) + (k+1)(k+5) = \frac 16*(k+1)(k+2)(2(k+1) + 13)$

$=\frac 16(k+1)(k+2)(2k + 15)$?

So does $\frac 16*k(k+1)(2k + 13) + (k+1)(k+5)=\frac 16(k+1)(k+2)(2k + 15)$?

Well, does it? Well expand those out and see.

If so you have proven it.

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Expand into factorial polynomials and sum the telescoping series. $$\begin{align}\sum_{k=1}^nk(k+4)&=\sum_{k=1}^n\left[k(k+1)+3k\right]\\ &=\sum_{k=1}^n\left[\frac13k(k+1)(k+2)+\frac32k(k+1)-\frac13(k-1)k(k+1)-\frac32k(k-1)\right]\\ &=\frac13n(n+1)(n+2)+\frac32n(n+1)-\frac13(0)(1)(2)-\frac32(1)(0)\\ &=\frac16n(n+1)(2n+4+9)=\frac16n(n+1)(2n+13)\end{align}$$

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