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If in the expansion of (1+x)n,a,b,c are three consecutive coefficients, then n =

a
ac+ab+bcb2+ac
b
2ac+ab+bcb2−ac
c
ab+acb2−ac
d
none of these

detailed solution

Correct option is B

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

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