If the coefficient of x^7 in the expansion of (ax^2 + 1 bx )¹¹ is equal to the coefficient of x⁻⁷ in the expansion of (ax - 1 b^2 x^2 )¹¹ , where a and b are positive real numbers, then the minimum value of a + 7b is

Step-by-step solution

Let the general term in the expansion of (ax^2 + 1 bx )¹¹ be T_ r+1 . T_ r+1 = ¹¹C_r (ax^2)^ 11-r ( 1 bx )^r = ¹¹C_r a^ 11-r b^ -r x^ 22-3r For the coefficient of x^7 , we set 22 - 3r = 7 3r = 15 r = 5 . Coefficient of x^7 = ¹¹C₅ a^6 b⁻⁵ . Now, let the general term in the expansion of (ax - 1 b^2 x^2 )¹¹ be T_ k+1 . T_ k+1 = ¹¹C_k (ax)^ 11-k (- 1 b^2 x^2 )^k = ¹¹C_k a^ 11-k (-1)^k b^ -2k x^ 11-3k For the coefficient of x⁻⁷ , we set 11 - 3k = -7 3k = 18 k = 6 . Coefficient of x⁻⁷ = ¹¹C₆ a^5 (-1)^6 b⁻¹² = ¹¹C₆ a^5 b⁻