Question

If the coefficient of $x^{2r}$ in the expansion of ${(x+\frac{1}{x^2})}^{n-3}$ is not zero, then $\left(\frac{n-2r}{3}\right)$ is:

• A. a rational number
• B. a positive integer
• C. a negative integer
• D. a positive rational number, but not an integer

Answer 1

Answer: (B)

a positive integer

Since the coefficient of $x^{2r}$ is not zero, there exists a term of $x^{2r}$ in the expansion of ${(x+\frac{1}{x^2})}^{n-3}$

Let the $r+1^{th}$ term be the term of $x^{2r}$

Hence,

$x^{-2r}\times x^{n-3-r}=x^{2r}$

Comparing the indices we get,

$n-3-r-2r=2r$

$n=5r+3$

Hence,

$\frac{n-2r}{3}=r+1$

This is a positive integer.