Question

Expand to 4 terms the following expressions :
$(4a-8x{)}^{-\frac{1}{2}}$

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,

$(4a-8x{)}^{-\frac{1}{2}}=(4a{)}^{-\frac{1}{2}}{(1-\frac{2x}{a})}^{-\frac{1}{2}}=\frac{1}{2\sqrt{a}}{(1-\frac{2x}{a})}^{-\frac{1}{2}}$

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

$=\frac{1}{2\sqrt{a}}[1+\frac{x}{a}+\frac{3}{2}.\frac{x^2}{a^2}+\frac{5}{2}.\frac{x^3}{a^3}+........]$