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If $$ (1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^r $$ then find the sum of : $$ a_0^2-a_1^2+a_2^2 +.....+(-1)^{n-1}a_{n-1}^2=\sum_{r=0}^{n-1}(-1)^{r-1}a_r^2 $$ in terms of $a_n$ and $n$.

What I've done : I've tried replacing $x$ with $\frac{-1}{x}$ and then multiplying the resultant series with the original series. However, this proved long and tedious and I gave up. Is there any other method to solve this question ? The answer given in my textbook is $\frac{a_n(1-(-1)^na_n)}{2}$. However I believe this answer to be wrong since the previous owner has written so in the margin.

Thankyou.

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Firstly note that the $a's$ have the following symmetry $a_{2n-r}=a_r$

\begin{eqnarray*} (1+x+x^2)^n =\sum_{r=0}^{2n} a_r x^{r} \end{eqnarray*} Now substitute $x=-x$ \begin{eqnarray*} (1-x+x^2)^n =\sum_{r=0}^{2n} (-1)^r a_r x^{r} \end{eqnarray*} Multiply these two equations \begin{eqnarray*} (1+x^2+x^4)^n =\cdots \end{eqnarray*} Now consider the coefficient of $x^{2n}$ ... we have \begin{eqnarray*} a_n = 2\sum_{r=0}^{2n} (-1)^r a_r^2 +(-1)^n a_n^2 \end{eqnarray*} So the answer in the book is right ... The answer is $\frac{a_n(1-(-1)^n a_n)}{2}$.

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    How do we prove that $a_{2n-r}=a_r$ ? – 2017-02-20
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    Substitute $x=1/x$ and multiply by $x^{2n}$. This will cause the series to reverse. – 2017-02-20
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    So we get the same series... Thanks. – 2017-02-20
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    You are welcome :-) – 2017-02-20
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    Try a few examples with small numbers $n$ and you will find that the answer in the book is correct and that this answer is wrong. E.g. $n=4$, $(a_r) = (1,4,10,16,19,16,10,4,1)$. You get $1^2-4^2+10^2-16^2 = -171 = \frac{19(1-19)}{2}$ – 2017-02-20
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    @ReinhardMeier ... goodness me, you are absolutely right ... I missed a square on the second $a_n$ ... I will edit now .... thank you. – 2017-02-20
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    Nicely done. (+1) – 2017-02-20
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    @Dr.MV Wow, cool ... Thank you. – 2017-02-20