Question

If $\left(1 + x-3x^{2}\right)^{10}= 1 + a_{1}x + a_{2}x^{2} + ... + a_{20}x^{20}$, then $a_{2} + a_{4}+a_{6} + ... + a_{20}$ is equal to

• A. $\frac{3^{10}+1}{2}$
• B. $\frac{3^{9}+1}{2}$
• C. $\frac{3^{10}-1}{2}$
• D. $\frac{3^{9}-1}{2}$

Answer 1

Answer: (C)

$\left(1 + x-3x^{2}\right)^{10}= 1 + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ... + a_{20}x^{20}$
Putting $x = 1$, we get
$\left(1 + 1 - 3\right)^{10} = 1 +a_{1} + a_{2}+a_{3} + ... + a_{20}$
$\therefore 1=1+a_{1} + a_{2}+a_{3} + ... + a_{20}\quad \ldots \left(i\right)$
Putting $x = - 1$, we get
$\left(1 - 1 - 3\right)^{10} = 1 -a_{1} + a_{2}-a_{3} + ... + a_{20}$
$\Rightarrow \left(3\right)^{10}=1-a_{1} + a_{2}-a_{3} + ... + a_{20}\quad \ldots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, $3^{10} + 1 = 2\left(1 + a_{2} + a_{4} + .....+ a_{20}\right)$
$\therefore a_{2} + a_{4}+a_{6} + ... + a_{20}= \frac{3^{10}+1-2}{2} = \frac{3^{10}-1}{2}$