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In the binomial expansion of [2^x + (1/4)^x]ⁿ (x is real, n is a positive integer), let T_r denote the r-th term. If T₃: T₂ = 7: 1 and the sum of the binomial coefficients of the second and third terms is 36, find n and x.

1. (A) n = 7
2. (B) n = 8
3. (C) x = 1/2
4. (D) x = -1/3

Correct answer: (B) n = 8

Solution

T₂ = C(n,1)*(2^x)^(n-1)*(1/4^x); T₃ = C(n,2)*(2^x)^(n-2)*(1/4^x)². Sum of binomial coefficients of 2nd and 3rd terms: C(n,1)+C(n,2) = n + n(n-1)/2 = n(n+1)/2 = 36 => n(n+1) = 72 => n = 8. T3/T2 = [C(8,2)/C(8,1)] * (1/4^x)/2^x = (28/8) * (1/4^x * 1/2^x) = (7/2) * 2^(-3x) = 7. So 2^(-3x) = 2 => -3x = 1 => x = -1/3. Correct options: B and D.

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