September 5th, 2019, 03:57 PM  #1 
Newbie Joined: Sep 2019 From: Dallas, TX Posts: 1 Thanks: 0  Help with a proof
Hey guys. I found out that a Σ2n n=1 EQUALS [a^2  a] However, although I can see that it works, I can't figure out why this is true. Can anyone figure this out? 
September 5th, 2019, 05:14 PM  #2 
Member Joined: Oct 2018 From: USA Posts: 93 Thanks: 66 Math Focus: Algebraic Geometry 
The reason that you may be having trouble is that the sum doesn't equal $a^2 a$, $a=1$ would imply $2 = 11 = 0$. The sum does equal $a^2 + a$ however. You can use mathematical induction, which is just proving the equality is true for the first term ($a=1$ in this case) and then assuming the equality is true for some $a$ and then proving the equality is true for $a+1$. This is basically just a domino type effect to prove it for all $a$ So, let $a=1$, then $2 = 1+1 = 2$ is true. Now, assume the equality is true for $a$, namely $\displaystyle \sum_{n=1}^{a}2n = a^2 + a$ We add $2(a+1)$ to both sides and go from there. $\displaystyle \sum_{n=1}^{a}2n + 2(a+1) = a^{2} +a + 2(a+1)$ $\displaystyle \sum_{n=1}^{a}2n + 2(a+1) = a^{2} +3a +2$ $\displaystyle \sum_{n=1}^{a}2n + 2(a+1) = (a+1)^2 + (a+1)$ and therefore $\displaystyle \sum_{n=1}^{a+1}2n = (a+1)^2 + (a+1)$ and the equality is proven. Last edited by Greens; September 5th, 2019 at 05:16 PM. Reason: Grammar 
September 5th, 2019, 07:52 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics 
\[ \sum_{n=1}^a 2n = \sum_{n=1}^a n + \sum_{n=1}^a a+1n = \sum_{n=1}^a a+1 = a(a+1). \] 
September 6th, 2019, 03:29 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
You can also predict that visually. Let a=5. You can split a 5x5 square into 5 (1x1) squares on the diagonal, plus a 1x4 and a 4x1 rectangle, plus a 1x3 and 3x1, plus a 1x2 and 2x1, plus two 1x1s. Add another 5, and you have 2 1s, 2 2s, 2 3s, 2 4s, and 2 5s.

September 6th, 2019, 06:46 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723 
You could use Gauss's (childhood) method. 2+4+....2a=2a+...+4+2. Add the two sequences term by term to get 2a+2 +(2a2)+4+....+2+2a=a(2a+2). Divide by 2 to get a(a+1).
