Question

Let ${(1+x+2x^2)}^{20}=a_0+a_{1}x+a_2{x}^2+\dots \dots +a_{40}{x}^{40}$. Then, $a_1+a_3+a_5+\dots .+a_{37}$ is equal to

• A. $2^{20}(2^{20}+21)$
• B. $2^{19}(2^{20}+21)$
• C. $2^{20}(2^{20}–21)$
• D. $2^{19}(2^{20}–21)$

Answer 1

The correct option is D

$2^{19}(2^{20}–21)$

Explanation for correct option

Determining the value of $a_1+a_3+a_5+\dots .+a_{37}$.

Given, ${(1+x+2x^2)}^{20}=a_0+a_{1}x+a_2{x}^2+\dots \dots +a_{40}{x}^{40}$

Putting $x=1$ we get,

$$\begin{gathered} {\left(1+1+2\right)}^{20}=a_0+a_1+....+a_{40} \\ \Rightarrow 4^{20}=a_0+a_1+....+a_{40}\left(1\right) \end{gathered}$$

Putting $x=-1$ in given expression we get,

${\left(2\right)}^{20}=a_0-a_1+a_2.....+a_{40}\left(2\right)$

Subtract equation $\left(2\right)$ from $\left(1\right)$ we get,

$$\begin{gathered} 4^{20}-2^{20}=2\left(a_1+a_3+a_5+......+a_{39}\right) \\ \Rightarrow \frac{4^{20}-2^{20}}{2}=\left(a_1+a_3+....+a_{39}\right) \\ \Rightarrow \left(a_1+a_3+....+a_{37}\right)=\frac{4^{20}-2^{20}}{2}-a_{39}\left(3\right) \end{gathered}$$

Now $a_{39}$ is the coefficient of $x^{39}$ in given expression. Hence, ${}^{20}C_{19}\times 2^{19}$ is the coefficient of $x^{39}$. Therefore,

$$\begin{gathered} a_{39}={}^{20}C_{19}\times 2^{19} \\ =20\times 2^{19} \end{gathered}$$

Put this value in equation $\left(3\right)$ we get,

$$\begin{gathered} a_1+a_3+....+a_{37}=\frac{4^{20}-2^{20}}{2}-\left(20\times 2^{19}\right) \\ =\frac{4^{20}-2^{20}-\left(20\times 2^{20}\right)}{2} \\ =\frac{2^{40}-2^{20}-\left(20\times 2^{20}\right)}{2} \\ =\frac{2^{20}\left(2^{20}-1-20\right)}{2} \\ =2^{19}\left(2^{20}-21\right) \end{gathered}$$

Therefore the value of $a_1+a_3+a_5.....+a_{37}$ is equal to $2^{19}\left(2^{20}-21\right)$

Hence, the correct answer is option (D).