How can I prove that$$\frac{(-1)^n(2n)!}{(n!)^2}+2\sum_{i=0}^{n-1}\frac{(-1)^i(2n)!}{i!(2n-i)!} = 0 $$ for every positive integer $n$?
I came across this identity when trying to prove that the inverse of an exponential matrix is the exponential matrix with the exponent negative, a.k.a. $[e^{\mathbf{A}t}]^{-1} = e^{-\mathbf{A}t} $
Or alternatively, could you help me sketch a simple proof of the mentioned matrix equation?
It’s a consequence of the binomial theorem:
$$\begin{align*} 0&=(1-1)^{2n}=\sum_{i=0}^{2n}(-1)^i\binom{2n}i\\ &=\sum_{i=0}^{n-1}(-1)^i\binom{2n}i+(-1)^n\binom{2n}n+\sum_{i=n+1}^{2n}(-1)^i\binom{2n}i\\ &=(-1)^n\binom{2n}n+\sum_{i=0}^{n-1}(-1)^i\binom{2n}i+\sum_{i=0}^{n-1}(-1)^{2n-i}\binom{2n}{2n-i}\\ &=(-1)^n\binom{2n}n+2\sum_{i=0}^{n-1}(-1)^i\binom{2n}i \end{align*}$$