Question

Find the $(r+1{)}^{th}$ term in each of the following expansion :
$\frac{1}{\sqrt{1+2x}}$

Answer 1

The general term i.e. $(r+1{)}^{th}$ term in the expansion of $(1+x{)}^n$ is given by

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

Now,

$\frac{1}{\sqrt{1+2x}}=(1+2x{)}^{-\frac{1}{2}}$

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

$=\frac{(-1)(-3)(-5)(-7)........(1-2r)}{2^r.r!}{2}^r.x^r$

$=(-1{)}^r\frac{\mathrm{1.3.5.7.......}(2r-1)}{r!}{x}^r$