Question

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

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,

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

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