Question

Question about 'The Sum of; 'Why is Step2 and Step3 equal to each other?

Answer 1

Okay?

Answer 2

You don't get this one?\[
\sum_{\Omega=1}^n\Omega =\frac{n(n+1)}{2}
\]

Answer 3

\[
\sum_{\Omega = 1}^{n} 1 =\underbrace{ 1+1+\ldots +1}_n = n
\]

Answer 4

Is the first one equal to each other? why is that?

Answer 5

is that a definition?

Answer 6

axiom? :/

Answer 7

So you want me to show you why this is true\[
\sum_{\Omega=1}^n\Omega =\frac{n(n+1)}{2}
\]is that correct?

Answer 8

Yes.

Answer 9

First of all \[
\sum_{\Omega=1}^n n-(\Omega-1) = n + (n-1) + \ldots + 1 = \sum_{\Omega=1}^n \Omega
\]Does that make sense?

Answer 10

Think about it \(\Omega = 1 \implies n-(\Omega - 1) = n\), and \(\Omega = n \implies n-(\Omega - 1) = 1\).
Basically it's the same sum, but with the terms in the same direction.

Answer 11

terms in opposite direction.

Answer 12

Ah, thank you I get it now.