Question

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. $3k+2$

Answer 1

The correct option is B

$3k$

Explanation for the correct option:

We know that $T_{r+1}={}^{n}c_r{x}^{n-r}{y}^r$ in the expansion of ${\left(x+y\right)}^n$.

So, in ${\left(x+\frac{2}{x^2}\right)}^n={\left(x+2x^{-2}\right)}^n$

$\begin{array}{rcl}{T}_{k+1}& =& {}^{n}c_k{x}^{n-k}{2}^k{x}^{-2k}\\ & =& {}^{n}c_k{x}^{n-3k}{2}^k\end{array}$

For $x^{2r}$

$\begin{array}{rcl}n-3k& =& 2r\\ & \Rightarrow & n-2r=3k\end{array}$

Hence, option B is correct.