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The expansion of (1+ax)n can be written using the binomial theorem as:

(1 + ax)^n = 1 + nax + \frac{n(n - 1)}{2!}(ax)^2 + \frac{n(n - 1)(n - 2)}{3!}(ax)^3 + \dotsb + (ax)^n.

The first few terms in this expansion are:

1 + nax + \frac{n(n - 1)}{2!}(ax)^2 + \frac{n(n - 1)(n - 2)}{3!}(ax)^3.

Simplifying the first three terms, we get:

1 - 6x + 12x^2 + cx^3.

We are given that the first three terms are:

1 - 6x + 12x^2

Comparing the coefficients of x3 on both sides of the equation, we get:

\frac{n(n - 1)(n - 2)}{3!} = c

To find c, we need to find (3n) and a. We can use the first two terms of the expansion to solve for a and n.

From the given first two terms, we have:

1 - 6x = 1 + nax

This implies that nax=20x or a=n20.

Substituting this value of a in the expression for the second term, we get:

nax = \frac{20}{n} \cdot x = 20x

Simplifying this equation gives:

n = 20

Therefore, n=20 and a=n20=1.

Now that we know n and a, we can find c:

c = \frac{n(n - 1)(n - 2)}{3!} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 4840

Therefore, the value of c is 4840.