Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$

Prove that $$ \frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n, \quad \forall k\in\mathbb{N}. $$ I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.

Solution 1:

If we define $$ f_n(t)=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{1} $$ then we have $$ \begin{align} f_n'(t) &=\sum_{k=0}^\infty\binom{2k+n}{k}kt^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n-1}{k-1}(2k+n)t^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n+1}{k}(2k+n+2)t^k\\[6pt] &=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)\tag{2} \end{align} $$

If we define $$ \begin{align} g_n(t) &=\frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n\\ &=\frac1{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^n\tag{3} \end{align} $$ then we have $$ g_n'(t) =\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\tag{4} $$ and therefore $$ \begin{align} &(n+2)g_{n+1}(t)+2tg_{n+1}'(t)\\[9pt] &=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &+2t\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+2}\\ &=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &+(1-\sqrt{1-4t})\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &=\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\[9pt] &=g_n'(t)\tag{5} \end{align} $$

Equations $(2)$ and $(5)$ ensure that $$ \begin{align} f_n'(t)&=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}f_{n+1}(t)\right]'\\ g_n'(t)&=(n+2)g_{n+1}(t)+2tg_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}g_{n+1}(t)\right]' \end{align}\tag{6} $$ Furthermore, it follows from $(1)$ and $(3)$ that $$ f_n(0)=g_n(0)=1\tag{7} $$

The generalized binomial theorem yields $$ \begin{align} (1-4t)^{-1/2} &=1+\frac124\frac{t}{1!}+\frac12\frac324^2\frac{t^2}{2!}+\frac12\frac32\frac524^3\frac{t^3}{3!}+\dots\\ &=\sum_{k=0}^\infty\frac{(2k-1)!!}{k!}2^kt^k\\ &=\sum_{k=0}^\infty\frac{(2k)!}{2^kk!k!}2^kt^k\\ &=\sum_{k=0}^\infty\binom{2k}{k}t^k\tag{8} \end{align} $$ Equation $(8)$ ensures that $f_0(t)=g_0(t)$.

Equations $(6)$, $(7)$, and $(8)$ ensure that $$ f_n(t)=g_n(t)\tag{9} $$ for all $n\ge0$. That is, $$ \frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n =\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{10} $$

Solution 2:

Due to a recent comment on my other answer, I took a second look at this question and tried to apply a double generating function. $$ \begin{align} &\sum_{n=0}^\infty\sum_{k=-n}^\infty\binom{2n+k}{n}x^ny^k\\ &=\sum_{n=0}^\infty\sum_{k=n}^\infty\binom{k}{n}\frac{x^n}{y^{2n}}y^k\\ &=\sum_{n=0}^\infty\frac{x^n}{y^{2n}}\frac{y^n}{(1-y)^{n+1}}\\ &=\frac1{1-y}\frac1{1-\frac{x}{y(1-y)}}\\ &=\frac{y}{y(1-y)-x}\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}-\frac{1-\sqrt{1-4x}}{1-\sqrt{1-4x}-2y}\right)\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}+\color{#C00000}{\frac{2x/y}{1+\sqrt{1-4x}-2x/y}}\right)\tag{1} \end{align} $$ The term in red contains those terms with negative powers of $y$. Eliminating those terms yields $$ \begin{align} \sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2n+k}{n}x^ny^k &=\frac1{\sqrt{1-4x}}\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{2y}{1+\sqrt{1-4x}}\right)^k\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{1-\sqrt{1-4x}}{2x}\right)^ky^k\tag{2} \end{align} $$ Equating identical powers of $y$ in $(2)$ shows that $$ \sum_{n=0}^\infty\binom{2n+k}{n}x^n=\frac1{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k\tag{3} $$