Show that $\sum\limits_{n=0}^{\infty} \binom{3n}{n} (\frac{4}{125})^n=\frac{1}{2\pi i}\int_{C(0;1)}\frac{-125}{(z-4)(4z^2+28z-1)}dz$

Question

Define $C(0;1)$ to be the circle with center $0$ and radius $1$ traversed in the counterclockwise direction. Show that $$\sum\limits_{n=0}^{\infty} \binom{3n}{n} \left(\frac{4}{125}\right)^n=\frac{1}{2\pi i}\int_{C(0;1)}\frac{-125}{(z-4)(4z^2+28z-1)}dz \tag{1}$$

I know that $$4z^2+28z-1=4\left(z-\frac{-7+5\sqrt 2}{2}\right)\left(z-\frac{-7-5\sqrt 2}{2}\right)$$ and I managed to show that $$\frac{1}{2\pi i}\int_{C(0;1)}\frac{-125}{(z-4)(4z^2+28z-1)}dz=\frac{15\sqrt 2}{28} +\frac{5}{14}$$ by Cauchy's Integral Formula where $z=\frac{-7+5\sqrt 2}{2}$ is the only singularity in $C(0;1)$.

Now my question is, is there a way to show $(1)$ by Laurent series expansion along with the binomial identity? I am clueless as to how I can proceed.

Answer 1

To deal with the LHS, start with the identity, valid for every integer $n\geqslant0$, $${3n\choose n}=[z^n](1+z)^{3n}=[z^0]z^{-n}(1+z)^{3n}$$ hence, for $|u|$ small enough, $$S(u)=\sum_{n=0}^\infty{3n\choose n}u^n$$ is $$S(u)=\frac1{2i\pi}\int_C\sum_{n=0}^\infty u^nz^{-n}(1+z)^{3n}\frac{dz}z=\frac1{2i\pi}\int_C\frac1{1-uz^{-1}(1+z)^3}\frac{dz}z$$ that is, $$S(u)=\frac1{2i\pi}\int_C\frac1{a(z)}dz$$ with $$a(z)=z-u(1+z)^3$$ For $u=4/125$, $a(z)$ has three simple poles, the only one inside $C$ is $$w=\frac{5\sqrt2-7}2$$ hence the residue formula yields $$S(u)=\frac1{a'(w)}=\frac1{1-3u(1+w)^2}=\frac{1+w}{1-2w}$$ and it remains to check that the last ratio is the desired value.