Question

If $a_0,a_1,a_2,\dots ,a_{2n}$ be the coefficients in the expansion of ${\left(1+x+x^2\right)}^n$ in ascending powers of $x$, then $a_0^2-a_1^2+a_2^2-a_3^2+\dots -a_{2n-1}^2+a_{2n}^2=$

• A. $a_{2n}$
• B. $a_n$
• C. $a_0$
• D. none of these

Answer 1

Answer: (B)

$\left(1+x+x^2\right)n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+$
$\dots +a_2{n}_{-1}{x}^2{n}^{-1}+a_{2}n{x}^{2}n\dots (1)$
Replacing $x$ by $\frac{1}{x}$ in (1), we get
$\left(1+x+x^2\right)n=a_0{x}^{2}n+a_1{x}^2{n}^{-1}+a_2{x}^2{n}^{-2}+\dots$
$+a_2{n}_{-1}x+a_{2}n$
Again, replacing $x$ by $-x$ in (1), we get
$\left(1-x+x^2\right)n=a_0-a_{1}x+a_2{x}^2-a_3{x}^3+\dots$
$-a_2{n}_{-1}{x}^2{n}^{-1}+a_{2}n{x}^{2}n\dots (2)$
Multiplying (1) and (2), we get
$\left(1+x^2+x^4\right)n=(a_0{x}^{2}n+a_1{x}^2{n}^{-1}+a_2{x}^2{n}^{-2}+$
$\dots +a_2{n}_{-1}x+a_{2}n)\times (a_0-a_{1}x+a_2{x}^2+$
$\dots -a_2{n}_{-1}{x}^2{n}^{-1}+a_{2}n{x}^{2}n)...$(3)
$[$ Note that $\left(1-x+x^2\right)\left(1+x+x^2\right)={\left(1+x^2\right)}^2-x^2$ $=1+x^2+x^4]$
Finally, replace $x$ by $x^2$ in (1), we get
$\left(1+x^2+x^4\right)n=a_0+a_1{x}^2+\dots +an{x}^{2}n+\dots +a_{2}n{x}^{4}n\dots (4)$
Now, equating the coefficients of $x^{2}n$ on the right hand sides of (3) and (4), we get
$a_0^2-a_1^2+a_2^2-a_3^2+\dots -a_{2n-1}^2+a_{2n}^2=an.$