To find the value of k, we need to understand the given expression and simplify it step by step.

Given:
(1 + x)^n = C0 + C1x + C2x^2 + ... + Cnx^n
(C1/C0) * (2C2/C1) * (3C3/C2) * ... * (nCn/Cn-1) = (1/k)^(n(n-1))

We can observe that the coefficient of x^r in the expansion of (1 + x)^n is given by nCr, where nCr represents the binomial coefficient.

To simplify the expression, let's consider each term in the product separately.

Term 1: (C1/C0)
The coefficient of x in the expansion of (1 + x)^n is nC1, so C1 = nC1 = n. Similarly, C0 = nC0 = 1.
Therefore, (C1/C0) = n/1 = n.

Term 2: (2C2/C1)
The coefficient of x^2 in the expansion of (1 + x)^n is nC2, so C2 = nC2. And we already know that C1 = n.
Therefore, (2C2/C1) = (2nC2)/(n) = 2nC2/n = 2n(n-1)/n = 2(n-1).

Similarly, we can simplify the other terms in the product.

Term 3: (3C3/C2) = 3(nC3)/(nC2) = 3(n(n-1)(n-2))/(n(n-1)) = 3(n-2).

Term 4: (4C4/C3) = 4(nC4)/(nC3) = 4(n(n-1)(n-2)(n-3))/(n(n-1)(n-2)) = 4(n-3).

And so on, until the last term:

Term n: (nCn)/(Cn-1) = (n(n-1)(n-2)...3*2*1)/(n(n-1)(n-2)...3*2) = n.

Now, we can rewrite the product as:

(n) * (2(n-1)) * (3(n-2)) * ... * (n) = (1/k)^(n(n-1))

Simplifying further:

n! = (1/k)^(n(n-1))

Taking the logarithm of both sides:

ln(n!) = ln[(1/k)^(n(n-1))]

Using the logarithmic property of exponents:

ln(n!) = (n(n-1)) * ln(1/k)

We know that ln(a/b) = ln(a) - ln(b), so:

ln(n!) = (n(n-1)) * [ln(1) - ln(k)]
ln(n!) = (n(n-1)) * (-ln(k))

Now, we have an equation in the form of y = mx, where y = ln(n!) and x = n(n-1).

This equation represents a straight line with slope m = -ln(k).

From the given options, we can see that the correct value of k is when m = 2, which corresponds to option B.

Hence,