已知 $\left(x + \dfrac{k}{x^2}\right)^n$ 是一个二项式展开式，其中 $k$ 和 $n$ 是正常数。

1. Apply the Binomial Theorem to expand the first 4 terms

$\left(x + \frac{k}{x^2}\right)^5 = x^5 + \binom{5}{1}x^4\left(\frac{k}{x^2}\right) + \binom{5}{2}x^3\left(\frac{k}{x^2}\right)^2 + \binom{5}{3}x^2\left(\frac{k}{x^2}\right)^3 + \dots$

2. Simplify each term

$= x^5 + 5x^4\left(\frac{k}{x^2}\right) + 10x^3\left(\frac{k^2}{x^4}\right) + 10x^2\left(\frac{k^3}{x^6}\right)$
$= x^5 + 5kx^2 + \frac{10k^2}{x} + \frac{10k^3}{x^4}$

1. Identify the relevant terms from the expansion of $\left(x + \frac{k}{x^2}\right)^5$

From part (a), the expansion is:
$\left(x + \frac{k}{x^2}\right)^5 = x^5 + 5kx^2 + 10k^2x^{-1} + 10k^3x^{-4} + \dots$

To find the term independent of $x$ in $\left(5x + \frac{3}{x^2}\right)\left(x + \frac{k}{x^2}\right)^5$, we need to multiply terms that result in $x^0$:

• $5x$ must be multiplied by the $x^{-1}$ term: $10k^2x^{-1}$
• $\frac{3}{x^2}$ must be multiplied by the $x^2$ term: $5kx^2$

2. Form the equation for the term independent of $x$

Term independent of $x$ = $(5x)(10k^2x^{-1}) + (3x^{-2})(5kx^2)$
$= 50k^2 + 15k$

Given that this term is 5:
$50k^2 + 15k = 5$

3. Solve the quadratic equation for $k$ and reject the invalid solution

$50k^2 + 15k - 5 = 0$
$10k^2 + 3k - 1 = 0$
$(5k - 1)(2k + 1) = 0$

$k = \frac{1}{5}$ or $k = -\frac{1}{2}$

Since $k$ is a positive constant, $k = \frac{1}{5}$.

1. Write down the general term of the expansion

The general term is given by:
$T_{r+1} = \binom{n}{r} (x)^{n-r} \left(\frac{k}{x^2}\right)^r$

2. Simplify the general term to find the power of $x$

$T_{r+1} = \binom{n}{r} x^{n-r} k^r x^{-2r}$
$T_{r+1} = \binom{n}{r} k^r x^{n-3r}$

3. Determine the condition for a term independent of $x$ and suggest two values for $n$

For a term independent of $x$, the power of $x$ must be zero:
$n - 3r = 0$
$n = 3r$
Since $r$ must be a non-negative integer, $n$ must be a multiple of 3. Given $n$ is a positive constant, possible values for $n$ are 3, 6, 9, etc.
Two possible powers of $n$ are 3 and 6.