Question

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

Answer 1

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

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

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

Now,

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

$=1+x+\frac{x^2}{6}-\frac{x^3}{54}+.....$