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?
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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 $ ... – Martin Sleziak Jan 16 '19 at 08:07
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... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there. – Martin Sleziak Jan 16 '19 at 08:08
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1From $\color{red}{\texttt{Mathematica}}$: $\frac{3}{4} \left(\text{HypergeometricPFQ}\left[, _3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{256}{27}\right)\right]-1\right)-\binom{4 n+3}{n+1} \text{HypergeometricPFQ}\left[, _4F_3\left(1,n+\frac{5}{4},n+\frac{3}{2},n+\frac{7}{4};n+\frac{4}{3},n+\frac{5}{3},n+2;\frac{256}{27}\right)\right]$ – Felix Marin Jan 17 '19 at 19:52
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1This 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 Feb 06 '19 at 10:07
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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
lab bhattacharjee
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