Question

Expand to 4 terms the following expressions :
${(1+\frac{x}{3})}^{-3}$

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,

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

$=1-x+\frac{2}{3}{x}^2-\frac{10}{27}{x}^3+.......$