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Suppose $n$ is a natural number and consider expansion: $$\left(1+x+x^2\right)^n=\sum_{r=0}^{2n} \ a_r x^r$$ Find $\ a_0+ \ a_3+ \ a_6+ \ a_9\ldots$

I used different method, but could not arrive at the answer.

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    $\begingroup$ Put $x=1,\omega,\omega^2$ and add $\endgroup$ Oct 25, 2017 at 7:11
  • $\begingroup$ Thanks got the answer $\endgroup$ Oct 25, 2017 at 7:18

2 Answers 2

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Note: I think the nice comment from @labbhattacharjee is worth an answer by its own. In order to obtain a formula for \begin{align*} a_0+a_3+a_6+\cdots \end{align*} with $(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^n$ it is convenient to consider the third roots of unity \begin{align*} 1,\,\omega_1=\frac{-1+i\sqrt{3}}{2},\,\omega_2=\frac{-1-i\sqrt{3}}{2} \end{align*} which are the zeros of \begin{align*} 1-x^3=(1-x)(1+x+x^2) \end{align*}

We observe that following holds \begin{align*} 1+\omega_1+\omega_1^2&=0\\ 1+\omega_2+\omega_2^2&=0\tag{1}\\ \omega_1^2&=\omega_2\\ 1&=\omega_1^3=\omega_2^3 \end{align*}

In order to obtain a formula for $a_0+a_3+a_6+\cdots$ we use the relationship (1) to filter out the coefficients $a_r$ with $r$ not a multiple of $3$. In order to do so, we evaluate the polynomial $(1+x+x^2)^{n}$ at $x=1,\omega_1$ and $\omega_2$.

We obtain \begin{align*} \color{blue}{3^n}&=(1+1+1)^n=a_0+a_1+a_2+a_3+a_4+a_5+a_6+\cdots\\ \color{blue}{0}&=(1+\omega_1+\omega_1^2)^n=a_0+a_1\omega_1+a_2\omega_1^2+a_3+a_4\omega_1+a_5\omega_1^2+a_6+\cdots\\ \color{blue}{0}&=(1+\omega_2+\omega_2^2)^n\\ &=(1+\omega_1^2+\omega_1)^n=a_0+a_1\omega_1^2+a_2\omega_1^1+a_3+a_4\omega_1^2+a_5\omega_1^1+a_6+\cdots \end{align*} Adding up these three equations we obtain \begin{align*} \color{blue}{3^n+0+0}&=3a_0+a_1(1+\omega_1+\omega_1^2)+a_2(1+\omega_1^2+\omega_1)\\ &\qquad +3a_3+a_4(1+\omega_1+\omega_1^2)+a_5(1+\omega_1^2+\omega_1)\\ &\qquad +3a_6+\cdots\\ &=3a_0+3a_3+3a_6+\cdots\\ &\color{blue}{=3(a_0+a_3+a_6+\cdots)} \end{align*}

We finally conclude \begin{align*} \color{blue}{a_0+a_3+a_6+\cdots=3^{n-1}\qquad\qquad n\geq 1} \end{align*}

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\begin{align} [1+x+x^2] =& \big[ (1+x)+x^2\big]^n \\ =& \sum_{k=0}^{n}\frac{n!}{k!(n-k)!}(1+x)^k\cdot (x^2)^{n-k} \\ =& \sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\big(\sum_{i=0}^k\frac{k!}{i!(k-i)!}x^i\big)\cdot (x^2)^{n-k} \\ =& \sum_{k=0}^{n}\sum_{i=0}^k\frac{n!}{(n-k)!(k-i)!i!}\cdot x^{2n-2k+i} \\ \end{align} Hire $\sum_{i=0}^{0}\frac{n!}{(n-k)!(k-i)!i!}=0$. This is the same as $$ \sum_{n\geq k\geq i\geq 0}\frac{n!}{(n-k)!(k-i)!i!}x^{2n-2k+i} $$ or to $v=n-k$, $u=k-i$ and $t=i$ we have $$ \sum_{r=0}^{n}\sum_{t+u+v=r}\frac{n!}{v!u!t!}x^{2v+t} $$

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