For some real number a and some positive integer n, the first few terms in the expansion of (1 + ax)^n

are 1 - 20x + 150x^2 + cx^3 +...

find c

Tacoflea Mar 20, 2023

#1**0 **

We know that the first few terms in the expansion of $(1+ax)^n$ are $1-20x+150x^2+cx^3+...$

Using the binomial theorem, we can expand $(1+ax)^n$ and get the first few terms: $$(1+ax)^n = 1 + nax + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + ...$$

Comparing the corresponding terms, we get: $$1=n(n-1)(n-2)(-a)^0/0!$$ $$-20=n(n-1)(n-2)(-a)^1/1!$$ $$150=n(n-1)(n-2)(-a)^2/2!$$ $$c=n(n-1)(n-2)(-a)^3/3!$$

Simplifying the second equation, we get: $$-20=n(n-1)(n-2)(-a)$$

Simplifying the third equation, we get: $$150=n(n-1)(-a)^2$$

Solving for $a$ in terms of $n$ from the second equation, and substituting it in the third equation, we get: $$150=n(n-1)\left(\frac{-20}{n(n-1)(n-2)}\right)^2$$

Solving for $n$, we get $n=6$.

Substituting this value of $n$ in the expressions for $a$ and $c$, we get: $$a = \frac{-20}{6\times5\times4} = -\frac{1}{3}$$ $$c = 6\times5\times4\left(-\frac{1}{3}\right)^3/3! = -40$$

Therefore, the value of $c$ is $-40$.

Note: We assumed that the ellipsis (...) denotes the additional terms in the expansion, and that the term $cx^3$ is the term with the coefficient of $x^3$. If this assumption is incorrect, the solution may be different.

Guest Mar 20, 2023