Question

Expand to 4 terms the following expressions :
$(2+x{)}^{-3}$

Answer 1

Whatever be value of $n$, negative integer or positive or negative fraction,

$(1+x{)}^n=1+\frac{n}{1!}x+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...+...+\frac{n(n-1)(n-2)...(n-r+1)}{r!}{x}^r+...+$ to $\infty$

where, $|x|<1$ i.e. $-1<x<1.$

Now,

$(2+x{)}^{-3}=2^{-3}{(1+\frac{x}{2})}^{-3}=\frac{1}{8}{(1+\frac{x}{2})}^{-3}$

$=\frac{1}{8}[1-3\times \frac{x}{2}+\frac{-3(-3-1)}{1.2}{\left(\frac{x}{2}\right)}^2+\frac{-3(-3-1)(-3-2)}{\mathrm{1.2.3}}{\left(\frac{x}{2}\right)}^3+...............]$

$=\frac{1}{8}[1-\frac{3}{2}x+\frac{3}{2}{x}^2-\frac{5}{4}{x}^3+........]$