If in the expansion of (1+x)n,a,b,c are three consecutive coefficients, then n =
ac+ab+bcb2+ac
2ac+ab+bcb2−ac
ab+acb2−ac
none of these
Here a=nCr,b=nCr+1 and c=nCr+2Put n = 2, r = 0, then option (b) holds the condition, i.e.,n=2ac+ab+bcb2−ac