Find the b and power value in the binomial with the given expansion of the first 3 terms

Question

In the expansion of $\left ( 1 + x \right )^{n}$, the first three terms are 1 - 0.9 + 0.36. Find the values of x and n.

I'm clueless as to how I would solve this......

Answer 1

The firts three terms are: $\quad(1+x)^n=1+nx+\dfrac{n(n-1)}2\,x^2=1-0.9+0.36$, so we have $$\begin{cases}nx=-0.9\\n(n-1)x^2=0.72\end{cases}$$ The second equation can be rewritten as $ $$(nx)^2-nx^2=0.81-nx^2=0.72$, so $nx^2=0.09$, from which we deduce $$x=\frac{nx^2}{nx}=-0.1,\;\text{ whence }\;n=9.$$

Answer 2

Generalizing Bernard's answer:

If $(1+x)^n = 1+a+b$, then $nx =a$ and $n(n-1)x^2/2 = b$.

Dividing the second equation by the first, $(n-1)x/2 = b/a$ or $(n-1)x = 2b/a$.

Subtracting this from the first, $x = a-2b/a =(a^2-2b)/a $ and $n = a/x =a/((a^2-2b)/a) =a^2/(a^2-2b) =1/(1-2b/a^2) $.

Check.

If $a=-.9$ and $b=.36$, then $x =-.9-(.72/-.9) =-.9+.8 =-.1$ and $n =a/x =-.9/(-.1) =9 $.