Q. If ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2n}{a_{r}x^{r}}$ then, $a_{1} - 2a_{2} + 3a_{3}\ldots - 2na_{2n}\$is equal to
A) $n$
B) $- n$
C) $0$
D) $2n$
Explanation
(b)
We have,
$(1 + x + x^{2})^{n} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} +$….$+ a_{2n}x^{2n}$
On differentiating both sides, we get
$n(1 - 1 + {1)}^{n - 1}(1 + 2x) = a_{1} + 2a_{2}x + 3a_{3}x^{2}$
+….$+ 2na_{2n}x^{2n - 1}$
On putting $x = - 1$ we get
$n(1 - 1 + 1)^{n - 1}(1 - 2) = a_{1} - 2a_{2} + 3a_{3} -$…$- 2na_{2n}$
$\Longrightarrow a_{1} - 2a_{2} + 3a_{3} -$…$- 2na_{2n} = - n$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.