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Summation Proof with Pascal's Triangle

My class was given the task to proof the equation in the picture.

I understand that it says any row, n, has a sum that is 1 less than the sum or all preceding rows, but alas I cannot proof it with an equation.

Any help is appreciated!

Attachment 25267

I am sorry for bad photo quality, I tried to highlight variables using MS Paint

Re: Summation Proof with Pascal's Triangle

Hi Hajmahkra,

I can't make out the image in your post, but based on your description of the problem, it should help to know that the sum of the elements in the nth row of Pascal's triangle is $\displaystyle 2^n$. The standard way to prove this, in case you don't already know, is to start with the binomial theorem,

$\displaystyle (1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i$

and then set x=1, with the result

$\displaystyle 2^n = \sum_{i=0}^n \binom{n}{i}$

That doesn't solve your problem completely, but it should be a good first step. Maybe you can do the rest.