Question

If x^2k occurs in the expansion of $\left( x + \frac{1}{x^{2}} \right)^{n–3}$, then

• A. n – 2k is a multiple of 2
• B. n – 2k is a multiple of 3
• C. k = 0
• D. None of these

Answer 1

Answer: (B)

n – 2k is a multiple of 3

Answer 2

(r + 1)th term in the expansion of $\left( x + \frac{1}{x^{2}} \right)^{n–3}$is given by T_r+1 = ^n–3C_r(x)^n–3–r $\left( \frac{1}{x^{2}} \right)^{r}$= ^n–3C_rx^n–3–3r

As x^2k occurs in the expansion of $\left( x + \frac{1}{x^{2}} \right)^{n–3}$,

we must have n –3 – 3r = 2k for some non-negative interger r.

Ž 3(1 + r) = n – 2k

Ž n – 2k is a multiple of 3