(1+x)n=1+nx+(n)(n−1)x22!+......(n!)xnn!
Where n is negative or fraction
given that
(1−4x)−12=1+(−12)x+(−12)(−12−1)x22!+......(−12)(−12−1)...(−12−r+1)xrr!
coefficient of xr is
(−1).(−3).(−5)....(1−2r)2r.r!
If the coefficients of xr−1,xr and xr+1 in the binomial expansion of (1+x)n are in AP, prove that n2−n(4r+1)+4r2−2=0