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:
- $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\ldots+a_{2 n} x^{2 n}$ $\quad$ …….(i)
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}} $ $\quad$ …….(ii)
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 $\left(1+t+t^{2}\right)^{n}=a_{n}$
