Question

If $(1+x+x^2{)}^{20}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40},$ then the value of $a_0^2-a_1^2+a_2^2-\cdots -a_{19}^2$ is

• A. $\frac{1}{2}{a}_{20}(1-a_{20})$
• B. $\frac{1}{2}{a}_{20}(1+a_{20})$
• C. $\frac{1}{2}{a}_{20}^2$
• D. \N

Answer 1

The correct option is A $\frac{1}{2}{a}_{20}(1-a_{20})$

Given
$(1+x+x^2{)}^{20}=\sum _{r=0}^{40}{a}_r{x}^r\cdots \left(1\right)$

Now, replacing $x$ by $\frac{1}{x}$, we get
${(1+\frac{1}{x}+\frac{1}{x^2})}^{20}=\sum _{r=0}^{40}{a}_r{\left(\frac{1}{x}\right)}^r\Rightarrow (1+x+x^2{)}^{20}=\sum _{r=0}^{40}{a}_r{x}^{40-r}\cdots \left(2\right)$

Comparing equation $\left(1\right)$ and $\left(2\right)$, we get
$a_0=a_{40}{a}_1=a_{39}$
So, $a_r=a_{40-r}$

Replacing $x$ by $-\frac{1}{x}$ in $\left(1\right)$, we get
${(1-\frac{1}{x}+\frac{1}{x^2})}^{20}=a_0-\frac{a_1}{x}+\frac{a_2}{x^2}-\cdots +\frac{a_{40}}{x^{40}}$
$\Rightarrow (1-x+x^{20}{)}^{20}=a_0{x}^{40}-a_1{x}^{39}+a_2{x}^{38}-\cdots +a_{40}$

Clearly,
$a_0^2-a_1^2+a_2^2-\cdots +a_{40}^2$
is the coefficient of $x^{40}$ in $(1+x+x^2{)}^{20}(1-x+x^2{)}^{20}$
$=$ Coefficient of $x^{40}$ in $(1+x^2+x^4{)}^{20}$
=coefficient of $y^{20}$ in $(1+y+y^2{)}^{20}$, where $y=x^2$
= $a_{20}$.

Hence, $a_0^2-a_1^2+a_2^2-\cdots +a_{40}^2=a_{20}$
$\Rightarrow (a_0^2-a_1^2+a_2^2-\cdots -a_{19}^2)+a_{20}^2$
$+(-a_{21}^2+a_{22}^2-a_{23}^2+\cdots +a_{40}^2)=a_{20}$
$\because a_r=a_{40-r}$, the above equation can be reduced to
$\Rightarrow 2(a_0^2-a_1^2+a_2^2-\cdots -a_{19}^2)+a_{20}^2=a_{20}$
$\Rightarrow a_0^2-a_1^2+a_2^2-\cdots -a_{19}^2=\frac{a_{20}}{2}(1-a_{20})$