Is the Summation of the Binomial Theorem Equal to Zero?

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The discussion centers on the Binomial Theorem, specifically the expression (x+y)^n and its summation form. When substituting x=1 and y=-1, the resulting sum, ∑_{k=0}^{n}{n choose k}(-1)^k, leads to the conclusion that it equals zero for all n except when n=0. In the case of n=0, the expression evaluates to 1, as 0^0 is conventionally defined as 1. The conversation highlights the nuances of binomial coefficients when n=0 and the implications for the theorem's summation. Overall, the summation of the Binomial Theorem equals zero for n greater than zero.
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Hello,

All we know the Binomial Theorm which may be stated mathematically as:

\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}y^k\,x^{n-k}

Now suppose that we have the following mathematical expression:

\sum_{k=0}^{n}{n\choose k}\,(-1)^k

if we substitute x=1 and y=-1 in the first equation we get the second. Is that mean the second equation is essentially zero, since (1-1)^n=0??

Regards
 
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Yes, indeed, unless n = 0.
 
Moo Of Doom said:
Yes, indeed, unless n = 0.

Why? In the case that n = 0, what will be the answer? 1?
 
00 is not well-defined and neither is 0Ck for any k <> 0 (although there are generalizations that extend the domain beyond the definition using just factorials).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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