Question

Expand to 4 terms the following expressions :
$(1+x{)}^{\frac{2}{5}}$

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,

$(1+x{)}^{\frac{2}{5}}=1+\frac{2}{5}x+\frac{\frac{2}{5}(\frac{2}{5}-1)}{1.2}{x}^2+\frac{\frac{2}{5}(\frac{2}{5}-1)(\frac{2}{5}-2)}{\mathrm{1.2.3}}{x}^3+.........$

$=1+\frac{2}{5}x-\frac{3}{25}{x}^2+\frac{8}{125}{x}^3+.....$