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,

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

$=1-x+2x^2-\frac{14}{3}{x}^3+.....$