Binomial Theorem 2 Question 13
16. If $n$ is a positive integer and
$$ \left(1+x+x^{2}\right)^{n}=a _0+a _1 x+\ldots+a _{2 n} x^{2 n} $$
Then, show that, $a _0^{2}-a _1^{2}+\ldots+a _{2 n}^{2}=a _n$.
(1994, 5M)
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Solution:
- $\left(1+x+x^{2}\right)^{n}=a _0+a _1 x+\ldots+a _{2 n} x^{2 n}$
Replacing $x$ by $-1 / x$, we get
$$ 1-\frac{1}{x}+{\frac{1}{x^{2}}}^{n}=a _0-\frac{a _1}{x}+\frac{a _2}{x^{2}}-\frac{a _3}{x^{3}}+\ldots+\frac{a _{2 n}}{x^{2 n}} $$
Now, $a _0^{2}-a _1^{2}+a _2^{2}-a _3^{2}+\ldots+a _{2 n}^{2}=$ coefficient of the term independent of $x$ in
$$ \begin{aligned} & {\left[a _0+a _1 x+a _2 x^{2}+\ldots+a _{2 n} x^{2 n}\right] } \\ & \times a _0-\frac{a _1}{x}+\frac{a _2}{x^{2}}-\ldots+\frac{a _{2 n}}{x^{2 n}} \end{aligned} $$
$=$ Coefficient of the term independent of $x$ in
$$ \left(1+x+x^{2}\right)^{n} 1-\frac{1}{x}+{\frac{1}{x^{2}}}^{n} $$
Now, RHS $=\left(1+x+x^{2}\right)^{n} 1-\frac{1}{x}+{\frac{1}{x^{2}}}^{n}$
$$ \begin{aligned} & =\frac{\left(1+x+x^{2}\right)^{n}\left(x^{2}-x+1\right)^{n}}{x^{2 n}}=\frac{\left[\left(x^{2}+1\right)^{2}-x^{2}\right]^{n}}{x^{2 n}} \\ & =\frac{\left(1+2 x^{2}+x^{4}-x^{2}\right)^{n}}{x^{2 n}}=\frac{\left(1+x^{2}+x^{4}\right)^{n}}{x^{2 n}} \end{aligned} $$
Thus, $\quad a _0^{2}-a _1^{2}+a _2^{2}-a _3^{2}+\ldots+a _{2 n}^{2}$
$=$ Coefficient of the term independent of $x$ in
$$ \frac{1}{x^{2 n}}\left(1+x^{2}+x^{4}\right)^{n} $$
$=$ Coefficient of $x^{2 n}$ in $\left(1+x^{2}+x^{4}\right)^{n}$
$=$ Coefficient of $t^{n}$ in $\left(1+t+t^{2}\right)^{n}=a _n$
