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September 5th, 2019, 03:57 PM   #1
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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?
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September 5th, 2019, 05:14 PM   #2
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The reason that you may be having trouble is that the sum doesn't equal $a^2 -a$, $a=1$ would imply $2 = 1-1 = 0$. 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
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September 5th, 2019, 07:52 PM   #3
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\[
\sum_{n=1}^a 2n = \sum_{n=1}^a n + \sum_{n=1}^a a+1-n = \sum_{n=1}^a a+1 = a(a+1).
\]
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September 6th, 2019, 03:29 AM   #4
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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.
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September 6th, 2019, 06:46 PM   #5
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You could use Gauss's (childhood) method. 2+4+....2a=2a+...+4+2. Add the two sequences term by term to get 2a+2 +(2a-2)+4+....+2+2a=a(2a+2). Divide by 2 to get a(a+1).
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