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 $a_0+3a_1+5a_2+\cdots +81a_{40}$ is:

• A. $161\times 3^{20}$
• B. $41\times 3^{40}$
• C. $41\times 3^{20}$
• D. $161\times 3^{40}$

Answer 1

The correct option is C $41\times 3^{20}$

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
$3^{20}=a_0+a_1+a_2+\cdots +a_{40}\Rightarrow 3^{20}=(a_0+a_1+a_2+\cdots +a_{19})+a_{20}+(a_{21}+a_{22}+a_{23}+\cdots +a_{40})\Rightarrow 3^{20}=2(a_0+a_1+a_2+\cdots +a_{19})+a_{20}(\because a_r=a_{40-r})\therefore a_0+a_1+a_2+\cdots ++a_{19}=\frac{1}{2}(9^{10}-a_{20})$

Also,
$a_0+3a_1+5a_2+\cdots +81a_{40}$
$=(a_0+81a_{40})+(3a_1+79a_{39})+\cdots +$
$(39a_{19}+43a_{21})+41a_{20}$
= $82(a_0+a_1+a_2+\cdots +a_{19})+41a_{20}$
=$41(9^{10}-a_{20})+41a_{20}$
$=41\times 3^{20}$

Alternate Solution:
$(1+x+x^2{)}^{20}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40}$
differentiating both the sides w.r.t. $x$
$\frac{d}{dx}(1+x+x^2{)}^{20}=a_1+2a_{2}x+3a_3{x}^2+\cdots +40a_{40}{x}^{39}\Rightarrow 20(1+x+x^2{)}^{19}\frac{d}{dx}(1+x+x^2)=a_1+2a_{2}x+3a_3{x}^2+\cdots +40a_{40}{x}^{39}\Rightarrow 20(1+2x)(1+x+x^2{)}^{19}=a_1+2a_{2}x+3a_3{x}^2+\cdots +40a_{40}{x}^{39}$
Now putting $x=1$, we get
$a_1+2a_2+3a_3+\cdots +40a_{40}=20\cdot 3\cdot 3^{19}=20\cdot 3^{20}$
Now
$a_0+3a_1+5a_2+\cdots +81a_{40}=a_0+a_1+a_2+\cdots +a_{40}+2(a_1+2a_2+3a_3+\cdots +40a_{40})=3^{20}+40\cdot 3^{20}=41\cdot 3^{20}$