Question

Find the general term in the expansion of $(1-nx{)}^{\frac{1}{n}}.$

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,

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

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

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

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

Since, $(-1{)}^r(-1{)}^{r-1}=(-1{)}^{2r-1}=-1$