Question

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

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,

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

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

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