Sum of product of binomial coefficients.

Question

Let $n\geq 1$ and define $$ S(n) = \binom{n}{0}\binom{n}{0} - \binom{n+1}{1}\binom{n}{1} + \binom{n+2}{2}\binom{n}{2} - \cdots + (-1)^n\binom{n+n}{n}\binom{n}{n} $$ In other words $$ S(n) = \sum_{i=0}^n (-1)^i\binom{n+i}{i}\binom{n}{i} $$

Then $S(n)=(-1)^n$

I am unable to make any progress.

Does this answer your question? Sum of product of binomial coefficients: $\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{n + k}{k} = (-1)^n$ – found with Approach0 — Martin R, Feb 21 '20 at 08:41

score 3 · Accepted Answer · answered Feb 21 '20 at 09:12

3

$$S_n=\sum_{k=0}^{n} (-1)^k {n+k \choose k} {n \choose k}=\sum_{k=0}^{n} (-1)^k {n+k \choose n} {n \choose k}.$$ $$\implies S_n= [x^n]\sum_{k=0}^{n} (-1)^k (1+x)^{n+k} {n \choose k}.$$ $$\implies S_n=[x^n] (1+x)^n(1-1-x)^n= (-1)^n,$$

answered Feb 21 '20 at 09:12

Z Ahmed

43,235

Sum of product of binomial coefficients.

1 Answers1