1. $\displaystyle \sum_{i\ =\ 1}^n i = \left(\begin{matrix}{n+1}\\{2}\end{matrix}\right)$

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- Jan 17th 2008, 10:08 AMEquinoXHelp me to prove or disprove this:
1. $\displaystyle \sum_{i\ =\ 1}^n i = \left(\begin{matrix}{n+1}\\{2}\end{matrix}\right)$

- Jan 17th 2008, 10:35 AMPlato
$\displaystyle {{ n + 1} \choose 2} = \frac{{(n + 1)!}}{{2!\left( {n - 1} \right)!}} = \frac{{\left( {n + 1} \right)(n)}}{2} = \sum\limits_{k = 1}^n k $

- Jan 17th 2008, 10:54 AMEquinoX
- Jan 17th 2008, 10:57 AMPlato
$\displaystyle \left( {n + 1} \right)! = \left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)!$

- Jan 17th 2008, 11:06 AMEquinoX
Oh I see, I got one more question to ask though:

Conjecture: The product of an even integer with an odd integer is even.

Proof(Direct): Let a be an even integer and let b be an odd integer. Because a is even, a = 2p for some integer p. Because b is odd, b = 2q + 1 for some integer q. If ab is even, then there exists an integer r such that ab = 2r. Thus, ab = (2p)(2q + 1) = 2r. As ab is equal to twice r, ab must be even.

Therefore, the product of an even integer with an odd integer is even

What is actually wrong with this proof??

My guess is because it tries to prove the other way around, it tries to proof if the product of two integer is even then one integer must be odd and the other must be even. Am I right?? - Jan 17th 2008, 05:41 PMJhevon
see rule 10 here

the problem with this proof is that it begs the question. that is, it assumes ab is even from the get-go. it doesn't prove this. the right way would be to expand 2p(2q + 1) and see if you can factorize it to the form 2r (here r is an expression that we can say is an integer, not a single variable)