Question

Find the general term in the expansion of $(1+x{)}^{\frac{1}{2}}$

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,

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

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

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

Since the number of factors in the numerator is $r$ and $r-1$ of these are negative, therefore by taking -1 out of each of these negative factors, we may write the above expression as

$=(-1{)}^{r-1}\frac{\mathrm{1.3.5......}(2r-3)}{2^r.r!}{x}^r$