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Algebra Level 3

Find $\displaystyle\sum_{k=0}^{2014}(-1)^k\left(\begin{matrix}2014\\k\end{matrix}\right)$ , where $\displaystyle\left(\begin{matrix}n\\r\end{matrix}\right)=\frac{n!}{(n-r)!r!}$ denotes combination .

For those of you who do not comprehend $\sum$ , find $\left(\begin{matrix}2014\\0\end{matrix}\right)-\left(\begin{matrix}2014\\1\end{matrix}\right)+\left(\begin{matrix}2014\\2\end{matrix}\right)-\left(\begin{matrix}2014\\3\end{matrix}\right)+\cdots-\left(\begin{matrix}2014\\2013\end{matrix}\right)+\left(\begin{matrix}2014\\2014\end{matrix}\right)$

The answer is 0.

3 solutions

Trevor Arashiro
Dec 2, 2014

For all n, $\displaystyle \sum^n_0 (-1)^x \dbinom{n}{x}=0$

This is the binomial expansion of $(a+b)^n=\displaystyle \sum^n_{x=0} (a^x )(b^{n-x})\dbinom{n}{x}$ where a=-1 and b=1

Jesus Ulises Avelar
Feb 25, 2015

Using the binomial theorem

Rajarshi Chatterjee
Dec 2, 2014

For every $n\geq r$ , $\dbinom{n}{r}$ (for all odd r)= $\dbinom{n}{r}$ (for all even r) .So, their difference is zero.

But 2014 is even!

Kenny Lau - 6 years, 6 months ago

the parity of r is required and not n.

Rajarshi Chatterjee - 6 years, 6 months ago

But since 2014 is even, the terms that are equal would not cancel out each other!

Kenny Lau - 6 years, 6 months ago

I guess what you are trying to say is that the sum of all odd r is equal to the sum of all even r? If so, please provide an explanation (since that is essentially the question).

What it looks like you are saying now, is that ${ n \choose 1 } = { n \choose 2}$ .

Calvin Lin Staff - 6 years, 5 months ago

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