Let n belongs to +ve integer and $$(1+x+x^2)^n=\sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0<r<2n}$$ as well as prove that $$\sum_{r=0}^{ n-1} a_r=\frac{1}{2}(3^n-a_n)$$. I tried to solve this problem by using multinomial theorem and comparing the coefficients but was confused how should I proceed please help me out.

  • $\begingroup$ Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal. $\endgroup$ – user10354138 Dec 6 '18 at 15:59
  • $\begingroup$ Should be $a_r = a_{2n-r}$. $\endgroup$ – gandalf61 Dec 6 '18 at 16:01

The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:

$(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}\sum_{r=0}^{2n} {a_rx^{-r}} = \sum_{r=0}^{2n} {a_rx^{2n-r}}=\sum_{r=0}^{2n} {a_{2n-r}x^r}$

$\Rightarrow \sum_{r=0}^{2n} {a_rx^r}=\sum_{r=0}^{2n} {a_{2n-r}x^r}$

Equating the coefficients of each power of $x$ gives the required result.

Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:

$(1+x+x^2)^n=\sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$

Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.


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