# Find the sum $S=\sum\limits_{k=1}^{n} {4k-1 \choose k}$

Find the sum $$S=\sum\limits_{k=1}^{n} {4k-1 \choose k}$$ I tried using the Pascal's identity to get $$S=\sum\limits_{k=1}^{n} {4k \choose k}-{4k-1 \choose k-1}$$ ,but it is not really telescopic. Any suggestions?

Hint:

The general term of $$(x^a+x^b)^{4k-1}$$ is $$\binom{4k-1}rx^{a(4k-1-r)}x^{br}$$

$$r=k\implies \binom{4k-1}k x^{a(3k-1)+bk}$$

To eliminate $$k$$ in the exponent of $$x$$

set $$3a+b=0\iff b=-3a$$

WLOG $$a=-1,b=?$$

We need to find the coefficient of $$x$$ in $$\sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence