+1364 For some real number $a$ and some positive integer $n,$ the first few terms in the expansion of $(1 + ax)^n$ are
\[1 - 6x + 12x^2 + cx^3 + \dotsb.\]
Find $c.$
+1768 1
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.
