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For the binomial expansion $ (1+x)^n = p_0 + p_1x+p_2x^2+...,$ where $p_i$ refer to the binomial coefficients.

Need to substitute the cube roots of unity (1, w, $w^2$) for the variable x; with w = $\dfrac{-1+i\sqrt{3}}{ 2},$ find the sums:\begin{align} (a) p_0 + p_3 + p_6 + ..., \\ (b) p_1 + p_4 + p_7 + ..., \\ (c) p_2 + p_5 + p_8 + ..., \\ \end{align} I have substituted the values as follows:

$(d)\;2^n = $$p_0+p_1+p_2+ p_3 + p_4 + p_5+...$

$(e)\;(1+w)^n = $$p_0+p_1.w+p_2.w^2+p_3+p_4.w+p_5.w^2+...$

$(f)\;(1+$$w^2$$)^n =(-w)^n = $$p_0+p_1.w^2+p_2.w+p_3+p_4.^2+p_5.w^2+...$

On adding (d), (e), (f); get the non-multiple of 3 terms vanished by the sum property of the cube roots of unity (1+w+$w^2$=0). Rest (non-zero) terms have each $p_i$ terms' coefficient being 3 (1+1+1=3) for all such terms.

(i) $(2^n + (1+w)^n + (-w)^n)/3 = p_0 + p_3 + p_6 + ...$

Geometrically/algebraically plotting the points (1+w), (1+$w^2$); we get: 1+w=$\dfrac{1+i\sqrt{3}}{2},$ 1+$w^2$=-w=$\dfrac{1-i\sqrt{3}}{2}.$

Obtaining the trigonometrical, & then exponential forms for the same, get:

(ii) 1+w=$\dfrac{1+i\sqrt{3}}{2}$= $cis(\dfrac{n\pi}{3})$= $e^\dfrac{in\pi}{3}$

(iii) -w =$\dfrac{1-i\sqrt{3}}{2}$= $cis(\dfrac{-n\pi}{3})$= $e^\dfrac{-in\pi}{3}$

Adding $e^\dfrac{in\pi}{3}$ + $e^\dfrac{-in\pi}{3}$ we get: 2.$\dfrac{e^\dfrac{in\pi}{3} + e^\dfrac{-in\pi}{3}}{2} = 2.\dfrac{cos(n\pi)}{3}$

Hence, get (i) modified as:

(iv) $p_0 + p_3 + p_6 + ... = \dfrac{2^n + 2cos(\dfrac{n\pi}{3})}{3}$

This solves only the part (a) of the question.

For the parts (b) & (c), there is no such closed form expression possible for me.

jiten
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  • The book has answer for part (b) as: $\dfrac{2^n - 2\cos((n+1)\dfrac{\pi}{3})}{3}; $

    and for part (c) as: $\dfrac{2^n - 2\cos((n-1)\dfrac{\pi}{3})}{3}$.

     – jiten Sep 20 '17 at 12:36
  • Hint: Instead of $(d)+(e)+(f)$, try computing $(d)+w\cdot(e)+w^2\cdot(f)$ and $(d)+w^2\cdot(e)+w\cdot(f)$... and rejoice. – Did Sep 20 '17 at 12:57
  • @Did : Thanks for that. But, if I could get the algebraic/geometric interpretation for the same, i.e. for multiplying by 1,w,$w^2$; and 1,$w^2$,w. Can we seek further permutations for multiplication factors - say multiply by $w^2$, w, 1; etc. – jiten Sep 20 '17 at 13:14
  • These are characters of the cyclic group generated by $w$. Since this group is isomorphic to $(\mathbb Z/3\mathbb Z,+)$, their linear span is the whole space of characters hence no other combination is needed. For example, $(w^2,w,1)$ is a multiple of $(1,w^2,w)$. – Did Sep 20 '17 at 13:19
  • @Did : Thanks for explanation. I understand that the 6 possible permutations are generated by the two multiplication groups : (1,w,$w^2$), (1,$w^2$,w). – jiten Sep 20 '17 at 22:54
  • @Did: But, would still like to ask how is $(w^2,w,1)$ a multiple of $(1,w^2,w)$. I know that both are permutations, but being a multiple is not clear. – jiten Sep 21 '17 at 12:13
  • $w^3=1$ hence $(w^2,w,1)=w^2\cdot(1,w^2,w)$. – Did Sep 21 '17 at 12:35
  • @Did: Thanks for that. But, I am still confused as to why any permutation (out of the 6 possible) is needed as a multiplication factor. The 3 series ((d), (e), (f)) have a default factor of 1. Why not choose a common factor of (w, w, w) or ($w^2, w^2, w^2$) instead? – jiten Sep 21 '17 at 21:48
  • Who said 6 permutations are needed? Please check the size of the group mentioned in my comments... – Did Sep 22 '17 at 04:06
  • @Did: Yes, size is 3, but there are 2 possible groups. I have partial answer to my question -(i) change the congruence class of the binomial coefficients,(ii) enable the removal of all binomial terms ($p_i$) not falling in congruence class mod 3 $\equiv$ 0. I say my answer is partial as it is not clear about algebraic/geometric interpretations. But, it is clear that for cancellation of binomial terms need have multiplication group from some permutation of either (1,w,$w^2$)/(1,$w^2$, w) to use-(1) property of sum of cube roots, (2)cube roots to form cyclic group on multiplication by w/$w^2$. – jiten Sep 22 '17 at 10:14

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Ill do part 2. Since all coefficients of $x^{1+3k}$ are to be retained, we multiply by $x^2$ to make the power $x^{3+3k}$, ie a multiple of $3$. $$(1+x)^n = a_o + a_1 x + a_2 x^2...a_nx^n$$

$$x^2(1+x)^n = a_ox^2+a_1x^3+a_2 x^4 ... a_nx^{n+2}$$

Now substituting cube roots of unity in above equation leaves required terms intact ($z^3 = 1$, where $z$ is cube root of unity).
All the other terms vanish, ( $1+z+z^2 =0$ if $z$ is cube root of unity)

Let $P(x) = x^2 (1+x)^n$. Then on we have:

$$\begin{align} a_1+a_4+a_7 ... &= \dfrac{P(1) + P(\omega) + P(\omega^2)}{3} \\ &= \dfrac{2^n + \omega^2(1+\omega)^n + \omega(1+\omega^2)^n}{3} \end{align}$$

jonsno
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  • Similar argument can be given for retaining the coefficients of $x^{2+3k}$, i.e. multiply by x . I am just extrapolating your statement, which means (as per my understanding) by $x^{1+3k}$, the terms: $p_1 +p_4 +p_7+...$ – jiten Sep 20 '17 at 13:23