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If n is a positive integer and three consecutive coefficients in the expansion of (1 + x)ⁿ are in the ratio 6: 33: 110, then the value of n is:

1. 9
2. 6
3. 12
4. 16

Correct answer: 12

Solution

With ratios C(n,r-1):C(n,r):C(n,r+1) = 6:33:110: (n-r+1)/r = 33/6 = 11/2 gives 2n+2 = 13r, and (n-r)/(r+1) = 110/33 = 10/3 gives 3n-10 = 13r. Equating: 2n+2 = 3n-10, so n = 12.

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