# Proving a nice pattern

Over on reddit, noncognitivism posted a nice sequence s/he had come across:

$4 + 1 = 5 = \sqrt{ (1)(2)(3)(4) + 1 }$ $(4+6) + 1 = 11 = \sqrt{ (2)(3)(4)(5) + 1}$ $(4+6+8) + 1 = 19 = \sqrt{ (3)(4)(5)(6) + 1}$ $…$ $(4+6+ … + (2k+2)) + 1 = \sqrt{ k(k+1)(k+2)(k+3) + 1 }$

Lovely. But why?

Well, the left hand side is an arithmetic series (with one added on) - in Core 1 terms, $a = 4$ and $d=2$, so its sum is $\frac{k}{2}\left[ 2(4) + (k-1)(2) \right] = \frac{k}{2} \left[ 8 + 2(k-1)\right] = k(k+3)$, nicely enough. So, we have that the left hand side is $k^2 + 3k + 1$.

The obvious - read, tedious - thing to do here would be to square both sides, multiply everything out and show we get two identical quartics. Yawn. And, strictly, not valid as a proof technique.

Instead, we’re going to do something clevererer, that involves no multiplying out at all: we’re going to gradually manipulate the left hand side into the same form as the right hand side. First up, it’s got a square root, so let’s throw one of those at it:

$k^2 + 3k + 1 = \sqrt{(k^2 + 3k + 1)^2}$

Then it ends with a +1. Now, we can’t just add one unless we subtract one too:

$k^2 + 3k + 1 = \sqrt{(k^2 + 3k + 1)^2-1 + 1}$

But oh, lookit, we’ve got the difference of two squares in there!

$k^2 + 3k + 1 = \sqrt{(k^2 + 3k)(k^2 + 3k + 2) + 1}$

And some quick factorising:

$k^2 + 3k + 1 = \sqrt{k(k+3)(k+2)(k+1) + 1}$

Oh, and let’s put it in the same order as we had to begin with:

$k^2 + 3k + 1 = \sqrt{k(k+1)(k+2)(k+3) + 1}~\blacksquare$