How find this sum $\sum_{j=0}^{\infty}\binom{m+2j}{m}t^{2j},0<t<1$

Question

Let $m$ is give postive integer numbers,

Find the sum $$\sum_{j=0}^{\infty}\binom{m+2j}{m}t^{2j},0<t<1$$

if this not have closed form,and can you use Special function ?

Answer 1

Since, for $-1<t<1$, we have $$\eqalign{ \sum_{k=0}^\infty\binom{m+k}{m}t^k&= \sum_{k=0}^\infty\frac{(m+k)\cdots (k+1)}{m!}t^k =\frac{1}{m!}\left(\sum_{k=0}^\infty t^{k+m}\right)^{(m)}\cr &=\frac{1}{m!}\left(\frac{1}{1-t}\right)^{(m)}=\frac{1}{(1-t)^{m+1}} }$$ we conclude that $$ \sum_{k=0}^\infty\binom{m+k}{m}t^k+\sum_{k=0}^\infty\binom{m+k}{m}(-t)^k=\frac{1}{(1-t)^{m+1}}+\frac{1}{(1+t)^{m+1}} $$ or $$ \sum_{j=0}^\infty\binom{m+2j}{m}t^{2j}=\frac{1}{2(1-t)^{m+1}}+\frac{1}{2(1+t)^{m+1}} $$