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How to get $$\binom n0 + \binom n3 + \binom n6 + \cdots$$

MY ATTEMPT

$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$

$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$

$$(1 + 1)^n = 2^n = \binom n0 + \binom n1 + \binom n2 + \cdots$$

$$(1+\omega)^n + (1+\omega^2)^n + (1 + 1)^n = 3 \left(\binom n0 + \binom n3 + \binom n6 + \cdots\right)$$

But how to solve LHS? I got the required equation in RHS

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marked as duplicate by user147263, user26857, Harish Chandra Rajpoot, Tim Raczkowski, drhab Dec 1 '15 at 8:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Your question isnt clear. $\endgroup$ – Archis Welankar Nov 26 '15 at 17:27
  • $\begingroup$ Two answer appear below, neither of them mine, and yet I'm the only person who's up-voted the question. $\endgroup$ – Michael Hardy Nov 26 '15 at 18:42
  • $\begingroup$ $\ldots\,$ and now three answers appear including mine and I'm still the only person who's up-voted the question. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 26 '15 at 18:56
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$$(1+\omega)^n+(1+\omega^2)^n+2^n\\ =(-\omega^2)^n+(-\omega)^n+2^n\\ =(-1)^n(\omega^{2n}+\omega^n)+2^n\\ $$ i) $n=3m$: $$(-1)^n(\omega^{2n}+\omega^n)+2^n=2\cdot(-1)^n+2^n$$ ii) $n=3m+1$ or $3m+2$: $$(-1)^n(\omega^{2n}+\omega^n)+2^n=(-1)^{n+1}+2^n$$

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  • $\begingroup$ The last equality does not always hold. Take $n=3$. $\endgroup$ – mathlove Nov 26 '15 at 17:43
  • $\begingroup$ @Kay K. what about LHD when $r=n$ $\endgroup$ – Mclalalala Mar 4 '17 at 9:59
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You have $$(1+\omega)^n+(1+\omega^2)^n+2^n=3\left(\binom n0+\binom n3+\binom n6+\cdots\right)$$

Now note that $$(1+\omega)^n+(1+\omega^2)^n=(-\omega^2)^n+(-\omega)^n=(-1)^n(\omega^n+\omega^{2n})$$

This is equal to $(-1)^n\cdot 2$ if $n\equiv 0\pmod 3$ or $(-1)^n\cdot (-1)=(-1)^{n+1}$ if $n\not\equiv 0\pmod 3.$

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  • $\begingroup$ I don't get the last line. LHS should just be (-1)^n(\omega^n+\omega^{2n}) right? I don't know what mod is. I think till here it is okay for me . $\endgroup$ – Ali Hasan Nov 27 '15 at 4:35
  • $\begingroup$ @AliHasan: $n\equiv 0\pmod 3$ means that $n$ is divisible by $3$. So, if $n$ is divisible by $3$, we have $\omega^n+\omega^{2n}=1+1=2$ because we have $\omega^3=1$. If $n$ is not divisible by $3$ (this is equivalent to $n\not\equiv 0\pmod 3$), then we have $\omega^n+\omega^{2n}=\omega+\omega^2=-1$ (why?) $\endgroup$ – mathlove Nov 27 '15 at 9:58
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Let $\omega = \dfrac{-1+ i\sqrt 3} 2 = \cos120^\circ + i\sin120^\circ$.

Then $\omega^3 = 1$, and $1+\omega = \cos60^\circ + i\sin60^\circ$, so $(1+\omega)^2 = \omega$.

A bit of arithmetic shows that $n\mapsto(1+\omega)^n + (1+\omega^2)^n$ is a periodic function with period $6$: \begin{array}{c|c} n & (1+\omega)^n + (1+\omega^2)^n \\ \hline 0 & \phantom{+}2 \\ 1 & \phantom{+}1 \\ 2 & -1 \\ 3 & -2 \\ 4 & -1 \\ 5 & \phantom{+}1 \end{array} Therefore $$ \binom n 0 + \binom n 3 + \binom n 6 + \cdots = \frac{2^n + \text{a periodic term never exceeding $2$ in absolute value} } 3. $$

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