- ( 1+x )
^{n}= 1+nx+[n( n-1)/2!] .x^{2}+ [n(n-1)(n-2)/3!].x^{3}+…… - ( 1+x )
^{-n}= 1-nx+[-n( n+1)/2!] .x^{2}– [n(n+1)(n+2)/3!].x^{3}+……

If x<<1
then x^{2},x^{3},…. is negligible. so:

- (1+x )
^{-n}≈ 1-nx - (1-x )
^{n}≈ 1-nx - (1-x )
^{-n}≈ 1+nx