Fundamental Drill#7

Expand the expression:

$\begin{aligned} \left(x^k + \frac 1x\right)^5 & = x^{5k} + 5x^{4k-1} + 10x^{3k-2} + 10x^{2k-3} + 5x^{k-4} + x^{-5} \end{aligned}$

The constant term exists when the power of $x$ is 0. For integer $k$ , there are only $\boxed{2}$ cases. When $k=0$ , then the first term $x^{5k} = x^0 = 1$ and when $x=4$ , then the second last term $5x^{k-4} = 5x^0 = 5$ .

$(x^k+\cfrac{1}{x})^5$

For $0\leq r\leq5$ ,

The $(r+1)th$ term in ascending order $=\dbinom{5}{r}(x^k)^{5-r}(\cfrac{1}{x})^r=\dbinom{5}{r}x^{k(5-r)-r}$

When $k(5-r)-r=0$

$\Rightarrow 5k-kr-r=0$

$5k-r(k+1)=0$

$r=\cfrac{5k}{k+1}$

$r=5+\cfrac{-5}{k+1}$

$0\leq r\leq5$

Upon checking for $r=0,1,2,3,4,5$ , we conclude that $k$ is integer only when $r=0,4$

So, there are $\boxed{2}$ possible values for $k$ .