Q. If ${(1 - x + x^{2})}^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$ , then the $a_{0} + a_{2} + a_{4} + ... + a_{2n}$ is equal to
A) $\frac{3^{n} + 1}{2}$
B) $\frac{3^{n} - 1}{2}$
C) $\frac{1 - 3^{n}}{2}$
D) $3^{n} + \frac{1}{2}$
Explanation
(a)
${(1 - x + x^{2})}^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$
On putting $x = 1$, we get
${(1 - 1 + 1)}^{n} = a_{0} + a_{1} + a_{2} + ... + a_{2n}$
$\Rightarrow 1 = a_{0} + a_{1} + a_{2} + ... + a_{2n}$ …(i)
Again, putting $x = - 1$, we get
$3^{n} = a_{0} - a_{1} + a_{2} - ... + a_{2n}$ …(ii)
On adding Eqs. (i) and (ii), we get
$\frac{3^{2} + 1}{2} = a_{0} + a_{2} + a_{4} + ... + a_{2n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.