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what is the value of $$\binom{n}{1}​+\binom{n}{4}+\binom{n}{7}​+\binom{n}{10}+\binom{n}{13}+\dots$$ in the form of number, cos, sin

attempts : I can calculate the value of $$\binom{n}{0}​+\binom{n}{3}+\binom{n}{6}​+\binom{n}{9}+\binom{n}{12}+\dots=\frac{1}{3}\left(2^n+2\cos \frac{n\pi}{3}\right)$$ by use primitive $3^\text{rd}$ root of the unity but this problem i cant solve it.

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  • $\begingroup$ Can you please show how you calculated the second sum? $\endgroup$ – RobPratt Apr 26 at 4:28
  • $\begingroup$ @RobPratt See AoPS or another post on this site for how to find the second sum. Found using Approach0. $\endgroup$ – Toby Mak Apr 26 at 4:33
  • $\begingroup$ @TobyMak, I know how to do both of them. I was looking for some effort from the OP. $\endgroup$ – RobPratt Apr 26 at 4:34
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Let $\omega=\exp(2\pi i/3)$. Then $$\frac{1+\omega^k+\omega^{2k}}{3}= \begin{cases} 1 &\text{if $3\mid k$}\\ 0 &\text{otherwise} \end{cases}$$ So \begin{align} \sum_{k=0}^\infty \binom{n}{3k+1} &=\sum_{k=0}^\infty \binom{n}{k+1}\frac{1+\omega^k+\omega^{2k}}{3} \\ &=\sum_{k=1}^\infty \binom{n}{k}\frac{1+\omega^{k-1}+\omega^{2(k-1)}}{3} \\ &=\frac{1}{3}\sum_{k=1}^\infty \binom{n}{k} + \frac{1}{3\omega}\sum_{k=1}^\infty \binom{n}{k}\omega^k + \frac{1}{3\omega^2}\sum_{k=1}^\infty \binom{n}{k} \omega^{2k} \\ &=\frac{1}{3}(2^n-1) + \frac{1}{3\omega}((1+\omega)^n-1) + \frac{1}{3\omega^2}((1+\omega^2)^n-1)\\ &=\frac{1}{3}(2^n-1) + \frac{\omega^2}{3}((1+\omega)^n-1) + \frac{\omega}{3}((1+\omega^2)^n-1)\\ &=\frac{2^n + \omega^2(1+\omega)^n + \omega(1+\omega^2)^n}{3} \end{align}

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As we are interested in every third term

Let $$\sum_{r=0}^{3r+1\le n}\binom n{3r+1}=\sum_{k=0}^2a_k(1+w_k)^n$$

where $w_k=w^k;k=0,1,2$ and $w$ is a complex cube root of unity so that $$1+w+w^2=0$$

Set $n=0,1,2$ to find $a_k;k=0,1,2$

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Binimial Series: $$(1+x)^n=\sum_{k=0}^{n} {n \choose k} x^k~~~~(1)$$ $w^3=1, 1+w+w^2=0$, let $x=1$ in (1) we get $$2^n=\sum_{k=0}^n {n \choose k}~~~(2)$$ Let $x=w$ in (1) and miltiply it by $w^2$, to get $$w^2(1+w)^n=(-1)^n w^{2n+2}=\sum_{k=1}^{n} w^{k+2} {n \choose k}~~~~~(3)$$ Let $x=w^2$ in (1) and multiply by $w$, to get $$w(1+w^2)^n=(-1)^n w^{n+1}=\sum_{k=0}^{n} w^{2k+1} {n \choose k}~~~~(4)$$ Now add (2), (3), (4), to get $$\sum_{k=0}^{n} [1+w^{k+1}+w^{2k+1}] {n \choose k}=2^n+(-1)^n[w^{2n+2}+w^{n+1}]$$ Whenever $k=3m+1$, $[1+w^{k+2}+w^{2k+1}]=[1+w^3+w^3]=3$, otherwise it vanishes as $[1+w+w^2]=0$ when $k\ne 2m+1$ So we get $$\sum_{m=0}^{n} {n\choose 3m+1}= \frac{1}{3}\left(2^n+(-1)^n[w^{2n+2}+w^{n+1}\right)=\frac{1}{3}(2^n+2\cos[(n-2)\pi/3])$$

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