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Have this sum have close form $$f(n)=\sum_{k=0}^{n-1}\left(\left(\sum_{i=0}^{k}(-1)^i\binom{n}{i}\right)\cdot\left(‌​\sum_{j=k+1}^{n}(-1)^j\binom{n}{j}\right)\right)$$

Maybe this sum can use integral to solve it? Thank you

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  • $\begingroup$ Note that David H answered first, I just added some extra justification. $\endgroup$ – robjohn Aug 8 '14 at 8:13
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I've simply added a justification of the first identity in David H's answer.

Assume that $n\ge1$, so that $\binom{n-1}{n}=0$. $$ \begin{align} \sum_{j=0}^k(-1)^j\binom{n}{j} &=\sum_{j=0}^k(-1)^j\left[\binom{n-1}{j}+\binom{n-1}{j-1}\right]\\ &=\sum_{j=0}^k(-1)^j\binom{n-1}{j}-\sum_{j=0}^{k-1}(-1)^j\binom{n-1}{j}\\ &=(-1)^k\binom{n-1}{k}\tag{1} \end{align} $$ Using $k=n$ in $(1)$ yields $$ \begin{align} \sum_{j=0}^n(-1)^j\binom{n}{j} &=(-1)^n\binom{n-1}{n}\\ &=0\tag{2} \end{align} $$

Therefore, using Vandermonde's Inequality and $(1)$ and $(2)$ gives $$ \begin{align} &\sum_{k=0}^n\left(\sum_{i=0}^k(-1)^i\binom{n}{i}\right)\left(\sum_{j=k+1}^n(-1)^j\binom{n}{j}\right)\\ &=\sum_{k=0}^n(-1)^k\binom{n-1}{k}(-1)^{k+1}\binom{n-1}{k}\\ &=-\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-1-k}\\ &=-\binom{2n-2}{n-1}\tag{3} \end{align} $$

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This sum is actually quite tame if you take it step by step.

The sum comprising the first factor is:

$$\sum_{j=0}^{k}(-1)^j\binom{n}{j}=(-1)^k\binom{n-1}{k}.$$

And since $\sum_{j=0}^{n}(-1)^j\binom{n}{j}=0$, we immediately find the sum in the second factor to be:

$$\sum_{j=k+1}^{n}(-1)^j\binom{n}{j}=-(-1)^k\binom{n-1}{k}.$$

Hence, the sum for $f(n)$ can be written as a single finite binomial sum whose values are well known to be the (negatives of the) central binomial coefficients:

$$f(n)=-\sum_{k=0}^{n-1}\binom{n-1}{k}^2=-\binom{2(n-1)}{n-1}.$$

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