Монгол Бодлогын Сан

Заавар: $$\sum\binom{2n}{n}z^n=\dfrac{1}{\sqrt{1+4z}}$$ $$\sum_{n}\binom{n+k}{k}x^n=\dfrac{1}{(1-x)^{k+1}}$$

Бодолт: \begin{align*} A(z)&=\sum_{n\ge 0} a_n z^n=\sum_{n\ge 0}\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}\binom{n+k}{k}z^n\\ &=\sum_{\mathstrut k}\sum\limits_{\mathstrut n}(-1)^k\binom{n}{k}\binom{n+k}{k}z^n=\sum_{\mathstrut k}\sum\limits_{\mathstrut n}(-1)^k\binom{2k}{k}\binom{n+k}{2k}z^n\\ &=\sum_{\mathstrut k}(-1)^k\binom{2k}{k}\sum\limits_{\mathstrut n}\binom{n+k}{2k}z^n=\sum_{\mathstrut k}(-z)^k\binom{2k}{k}\sum\limits_{\mathstrut n}\binom{n+2k}{2k}z^{n}\\ &=\sum_{\mathstrut k}\binom{2k}{k}\dfrac{(-z)^k}{(1-z)^{2k+1}}=\dfrac{1}{1-z}\sum_{\mathstrut k}\binom{2k}{k}\left(\dfrac{-z}{(1-z)^{2}}\right)^k\\ &=\dfrac{1}{1-z}\cdot\dfrac{1}{\sqrt{1-4\cdot\dfrac{-z}{(1-z)^2}}}=\dfrac{1}{\sqrt{(1-z)^2+4z}}=\dfrac{1}{1+z}\\ &=1-z+z^2-z^3+\dots \end{align*} тул $a_n=(-1)^n$ болж батлагдав.