Question

Expand to 4 terms the following expressions :
$(9+2x{)}^{\frac{1}{2}}$

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,

$(9+2x{)}^{\frac{1}{2}}=9^{\frac{1}{2}}{(1+\frac{2x}{9})}^{\frac{1}{2}}=3{(1+\frac{2x}{9})}^{\frac{1}{2}}$

$=3[1+\frac{1}{2}\times \frac{2x}{9}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{1.2}{\left(\frac{2x}{9}\right)}^2+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{\mathrm{1.2.3}}{\left(\frac{2x}{9}\right)}^3+...............]$

$=3[1+\frac{x}{9}-\frac{x^2}{162}+\frac{x^3}{1458}+........]$