Question

If $a_{0}, a_{1}, a_{2}, \ldots, a_{2 n }$ be the coefficients in the expansion of $\left(1+x+x^{2}\right)^{ n }$ in ascending powers of $x$, then $a_{0}^{2}-a_{1}^{2}+a_{2}^{2}-a_{3}^{2}+\ldots-a_{2 n-1}^{2}+a_{2 n}^{2}=$

• A. $a_{2 n }$
• B. $a_{ n }$
• C. $a_{0}$
• D. none of these

Answer 1

Answer: (B)

$\left(1+x+x^{2}\right) n=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+ $
$\ldots+a_{2} n_{-1} x^{2} n^{-1}+a_{2} n x^{2} n \ldots(1)$
Replacing $x$ by $\frac{1}{x}$ in (1), we get
$\left(1+x+x^{2}\right) n=a_{0} x^{2} n+a_{1} x^{2} n^{-1}+a_{2} x^{2} n^{-2}+\ldots $
$+a_{2} n_{-1} x+a_{2} n$
Again, replacing $x$ by $-x$ in (1), we get
$\left(1-x+x^{2}\right) n=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+\ldots $
$-a_{2} n_{-1} x^{2} n^{-1}+a_{2} n x^{2} n \ldots(2)$
Multiplying (1) and (2), we get
$\left(1+x^{2}+x^{4}\right) n=\left(a_{0} x^{2} n+a_{1} x^{2} n^{-1}+a_{2} x^{2} n^{-2}+\right. $
$\left.\ldots+a_{2} n_{-1} x+a_{2} n\right) \times\left(a_{0}-a_{1} x+a_{2} x^{2}+\right. $
$\left.\ldots-a_{2} n_{-1} x^{2} n^{-1}+a_{2} n x^{2} n\right)...$(3)
$\left[\right.$ Note that $\left(1-x+x^{2}\right)\left(1+x+x^{2}\right)=\left(1+x^{2}\right)^{2}-x^{2}$ $\left.=1+x^{2}+x^{4}\right]$
Finally, replace $x$ by $x^{2}$ in (1), we get
$\left(1+x^{2}+x^{4}\right) n=a_{0}+a_{1} x^{2}+\ldots+a n x^{2} n+\ldots+a_{2} n x^{4} n \ldots(4)$
Now, equating the coefficients of $x^{2} n$ on the right hand sides of (3) and (4), we get
$a_{0}^{2}-a_{1}^{2}+a_{2}^{2}-a_{3}^{2}+\ldots-a_{2 n-1}^{2}+a_{2 n}^{2}=a n .$