Value of sum of binomials: $P = \binom{N}{0}-\binom{N}{1}+\binom{N}{2}-\binom{N}{3}+ \dotsb + (-1)^N\binom{N}{N}$

Question

$P = \binom{N}{0}-\binom{N}{1}+\binom{N}{2}-\binom{N}{3}+ \dotsb + (-1)^N\binom{N}{N}$

I can calculate the value of this equation manually, but there any direct formula for calculating the value of this equation? Can anyone please help?

What is $P$ and what is $C(a,b)$? I presume from your tag that they represent permutations and combinations? You can't assume people know what you're talking about just because you do. — Edward Evans, Mar 31 '16 at 13:26
So $P$ just represents the value of that sum? That is,
$$P = \sum_{i = 0}^N C(N, i)$$ — Edward Evans, Mar 31 '16 at 13:31

score 5 · Answer 1 · answered Mar 31 '16 at 13:35

5

Assuming you mean $C(n,k)=\binom{n}{k}=\frac{n!}{k!(n-k)!}$, then you now from the binomial theorem that $$(a+b)^n=\sum_{k=0}^{n} \binom{n}{k} a^kb^{n-k}.$$ Now set $a=1$ and $b=-1$.

answered Mar 31 '16 at 13:35

Elmar Zander

1,625

Value of sum of binomials: $P = \binom{N}{0}-\binom{N}{1}+\binom{N}{2}-\binom{N}{3}+ \dotsb + (-1)^N\binom{N}{N}$

1 Answers1