Q. $If\ x^{2r}\ occurs\ in\ \left( x + \frac{2}{x^{2}} \right)^{n},\ then\ n - 2r\ must\ be\ of\ the\ form$

A) $3k - 1$
B) $3k$
C) $3k + 1$
D) $3k + 2$
Explanation

(b)
$The\ general\ termof\ \left( x + \frac{2}{x^{2}} \right)^{n}is$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ T}_{R + 1} = \ ^{n}C_{R}(x)^{n - R}\left( \frac{2}{x^{2}} \right)^{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ ^{n}C_{R}x^{n - 3R}2^{R}$
For $x^{2r}$ occurs, it means
${\ \ \ \ \ \ \ \ \ \ \ \ \ n - 3R = 2r
}{\Longrightarrow \ \ \ \ \ \ \ n - 2r = 3R}$
Hence, $n - 2r$ is of the form $3k$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.