By using generating functions and snake-oil I got to
Also what is the implication of $\sum \limits_ {k<={n}}$?
I am told that this is equivalent to:
But I'm not sure how to do that step, thanks for the help!
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4Does this answer your question? $\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n>m\geq 0$ – Maximilian Janisch Dec 19 '19 at 01:05
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It’s the binomial theorem,
$$(a+b)^n=\sum_{k\le n}\binom{n}ka^kb^{n-k}\;.$$
Now let $a=1+x$ and $b=-1$.
In this case taking the sum over $k\le n$ is equivalent to taking it from $k=0$ to $k=n$: by definition $\binom{n}k=0$ if $k<0$.
Brian M. Scott
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thanks a lot! I came across a similar question $\sum \limits_{k} {n \choose k} {x^k} {(1+y)^k}$ how would i apply the binomial theorem to this question. – wallace tang May 10 '15 at 23:26
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1@wallacetang: You’re welcome. Now that you’ve edited it, that’s clearly $\big(1+x(1+y)\big)^n$ by the binomial theorem, with $a=x(1+y)$ and $b=1$. – Brian M. Scott May 10 '15 at 23:29
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it comes out to be $(1+x+xy)^n$, im not sure how the xy term came to be – wallace tang May 10 '15 at 23:31