if x<12,then the coefficient of xr in the expansion of 1+2x1-2x2 is.
r2r
2r-12r
r22r+1
2r+12r
Explanation for the correct answer:
Find the value of xr:
Given, 1+2x1-2x2
We know the general formula for the expansion 1+xnis
Tr+1=nn-1.....n-r+1r!xr
So, the given equation can be written as,
1+2x1-2x2=1+2x1-2x-2=1+2x1+-2-2x+-2-2-12!-2x2+....+-2-2-1-2-2.....-2-r-1+1r-1!-2xr-1+-2-2-1-2-2.....-2-r+1r!-2xr=1-2-2-1-2-2.....-2-r+1r!-2xr+2x-2-2-1-2-2.....-2-r-1+1r-1!-2xr-1=xrr+12r+2.r.2r-1[∵r!=r(r-1)(r-2)(r-3).....]=xrr+12r+r.2r=xr2r+1.2r
Hence, the correct option is D.