A little help with a binomial theorem proof

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SUMMARY

The discussion centers on proving the identity (n,0) - (n,1) + ... + (-1)^n(n,n) = 0 using the binomial theorem. The proof is straightforward for odd n, as terms cancel each other out. However, for even n, the middle term alternates signs, complicating the proof. The solution involves applying the binomial theorem, specifically using the expression (x+y)^n = ∑(n,i)x^iy^(n-i) with appropriate values for x and y, particularly utilizing the fact that 1^i = 1 for all i.

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  • Basic algebraic manipulation skills
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Homework Statement


(here, (n,k) reads n choose k)
prove that (n,0) - (n, 1) + ... + (-1)n(n,n) = 0


Homework Equations



binomial theorem

The Attempt at a Solution


so this proof is relatively straightforward when n is odd. it's just matching up terms and having them cancel each other out. I'm having a problem proving it when n is even, because each term doesn't match up exactly. and the middle term also alternates between plus or minus depending on whether n/2 is even. (i think i have the middle term is (-1)n(n,n/2).
but anyway, I've been having trouble with it. a little hint or two would be nice. gracias!
 
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That's a good approach, but fails for even n, like you noticed.

Consider directly applying the binomial theorem

(x+y)^n = \sum_{i=0}^n (n,i)*x^iy^{n-i}

Now, just pick the right values for x,y.

Remember the obvious fact that 1^i = 1,\;\forall i.
 
<slaps forehead>
this is exactly the same as the proof i did before, except for the different x and y values.
thanks.
 

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