Question

Expand to 4 terms the following expressions :
$(1+x{)}^{\frac{3}{2}}$

Answer 1

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

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

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

Now,

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

$=1+\frac{3}{2}x+\frac{3}{8}{x}^2-\frac{1}{16}{x}^3+.....$