So, I know that, $\sum_{k=0}^n(-1)^k*\binom{n}{k} = 0$. I know the reason for this is because the summation of even subsets is equal to the summation of odd subsets and obviously $n=n \implies n-n=0$. However, I want to know 1)what and how to derive the formula for this using the definition of ($n$ choose $k$)and 2)Why the formula is what it is. I thought that maybe the formula is $(-2)^n$ since the formula for the non-alternating row sums is $2^n$ but I feel like that logic may be flawed.
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1Which definition of $\binom nk$ is "*the*" definition that you want to use? – bof Feb 20 '17 at 09:49
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Instead of considering that $(2)^N$ has to be replaced by $(-2)^N$...
Consider that $(1+1)^N$ is replaced by $(1-1)^N=0$ (the desired RHS !), and then use binomial expansion for the LHS.
Jean Marie
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