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 $a_0+a_1+a_2+\cdots +a_{19}$ is

• A. $\frac{1}{2}(9^{10}+a_{20})$
• B. $\frac{1}{2}(9^{10}-a_{20})$
• C. $\frac{9^{10}}{2}$
• D. $\frac{1}{2}(a_{20}-9^{10})$

Answer 1

The correct option is B $\frac{1}{2}(9^{10}-a_{20})$

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, $a_r=a_{40-r}$

Putting $x=1$ in equation $\left(1\right)$, we get
$3^{20}=a_0+a_1+a_2+\cdots +a_{40}\Rightarrow 3^{20}=(a_0+a_1+a_2+\cdots +a_{19})+a_{20}+(a_{21}+a_{22}+a_{23}+\cdots +a_{40})\Rightarrow 3^{20}=2(a_0+a_1+a_2+\cdots +a_{19})+a_{20}(\because a_r=a_{40-r})\therefore a_0+a_1+a_2+\cdots ++a_{19}=\frac{1}{2}(9^{10}-a_{20})$