# Let n belongs to +ve integer and $(1+x+x^2)^n=\sum_{r=0}^{2n} {a_rx^r}$ prove that: $a_r=a_{0<r<2n}$

Let n belongs to +ve integer and $$(1+x+x^2)^n=\sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0 as well as prove that $$\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.

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