Question

If there is a term containing $x^{2r}$ in ${(x+\frac{1}{x^2})}^{n-3}$, then

• A. n - 2r is a positive integral multiple of 3.
• B. n - 2r is even
• C. n - 2r is odd
• D. None of the above

Answer 1

Answer: (A)

n - 2r is a positive integral multiple of 3.

Suppose $(s+1{)}^{th}$ term conains $x^{2r}$
Then, we have
$T_{s+1}{=}^{n-3}{C}_s{x}^{n-3-s}{\left(\frac{1}{x^2}\right)}^s$
${=}^{n-3}{C}_s{x}^{n-3-3s}$
This will contain $x^{2r}$, if
$n-3-3s=2r$
$\Rightarrow s=\frac{n-3-2r}{3}$
$\Rightarrow s=\frac{n-2r}{3}-1$
$\Rightarrow s+1=\frac{n-2r}{3}$
$\Rightarrow n-2r=3(s+1)$
Hence, $(n-2r)$ is a positive integral multiple of $3$.