Dot product of taylor series $\sqrt{1+x}$

Question

I have to prove that $$ \sum_{k=1}^n \alpha_k \cdot \alpha_{n-k+1} = 0, $$ where $n>2$ and $\alpha_k$ is the k-th member in taylor series of $\sqrt{1+x}$. Namely, $$ \alpha_k = \frac{(-1)^k(2k)!}{(1-2k)(k!)^2(4^k)}x^k. $$ How can I do it? I tried to use a mathematical induction, but I didn't achieved any result.

Answer 1

Cauchy product is what you look for.

$\beta_{n-1} = \sum_{k=0}^{n-1} a_k a_{n-k-1}$ is the coefficient in the series developpement of $$ \left( \sqrt{1+x} \right)^2 = \sum_{n=0}^\infty \beta_n x^n $$ around 0, that is 0 when $n>2$.