# 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?

• You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots$, where $n$ is a positive integer., Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots$ ... Commented Jan 16, 2019 at 8:07
• Commented Jan 16, 2019 at 8:08
• From $\color{red}{\texttt{Mathematica}}$: $\frac{3}{4} \left(\text{HypergeometricPFQ}\left[\, _3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{256}{27}\right)\right]-1\right)-\binom{4 n+3}{n+1} \text{HypergeometricPFQ}\left[\, _4F_3\left(1,n+\frac{5}{4},n+\frac{3}{2},n+\frac{7}{4};n+\frac{4}{3},n+\frac{5}{3},n+2;\frac{256}{27}\right)\right]$ Commented Jan 17, 2019 at 19:52
• This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
– Did
Commented Feb 6, 2019 at 10:07

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