To find the values of $n$ and $k$ from the given binomial expansion of $(1+kx)^n$, we can compare the coefficients of the expansion with those provided in the problem.

The binomial expansion of $(1+kx)^n$ is given by:

$(1 + kx)^n = 1 + \binom{n}{1}kx + \binom{n}{2}k^2x^2 + \cdots + \binom{n}{n}k^n x^n$

Comparing this with the given expansion $1 + \frac{9x}{2} + 15k^2x^2 + \cdots + k^n x^n$:

1. For the coefficient of $x$:

   $\binom{n}{1}k = \frac{9}{2}$

   Simplifying, we have: nk = \frac{9}{2} \quad \text{(since \binom{n}{1} = n)}

2. For the coefficient of $x^2$:

   $\binom{n}{2}k^2 = 15k^2$

   Simplifying, we have: $\frac{n(n-1)}{2} = 15$ $n(n-1) = 30$

   This is a quadratic equation: $n^2 - n - 30 = 0$

   Solving this quadratic equation using the quadratic formula: $n = \frac{-(-1) \pm \sqrt{(-1)^2 + 4(1)(30)}}{2(1)}$ $n = \frac{1 \pm \sqrt{1 + 120}}{2}$ $n = \frac{1 \pm \sqrt{121}}{2}$ $n = \frac{1 \pm 11}{2}$

   So, $n = 6$ or $n = -5$. Since $n$ must be a positive integer, we have $n = 6$.

3. Substituting $n = 6$ back into the equation $nk = \frac{9}{2}$:

   $6k = \frac{9}{2}$ $k = \frac{9}{12} = \frac{3}{4}$

Thus, the values of $n$ and $k$ are:

$n = 6, \quad k = \frac{3}{4}$

Would you like further details on the binomial expansion, or do you have any other questions?

Here are 5 related questions:

1. How do you derive the general term in a binomial expansion?
2. What is the significance of the binomial coefficient $\binom{n}{r}$?
3. How can the binomial theorem be applied to approximate functions?
4. What happens to the binomial expansion when $k = 1$?
5. Can you use the binomial expansion to calculate probabilities in a binomial distribution?

Tip: When solving quadratic equations, always check the discriminant ($\Delta = b^2 - 4ac$) first to determine the nature of the roots.