Q. If ${(1 - x + x^{2})}^{n} = a_{0} + a_{1}x +$…..$a_{2n}x^{2n}$ then the value of $a_{0} + a_{2} + a_{4} +$….$+ a_{2n}$ is
A) $3^{n} + \frac{1}{2}$
B) $3^{n} - \frac{1}{2}$
C) $\frac{3^{n} - 1}{2}$
D) $\frac{3^{n} + 1}{2}$
Explanation
(d)
${(1 - x + x}^{2})^{n} = a_{0} + a_{1}x +$…$+ a_{2n}x^{2n}$
Putting $x = - 1$ and $1$
Successively and adding, we get
$a_{0} + a_{2} + a_{4} + ... + a_{2n} = \frac{3^{n} + 1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.