If the sum of the coefficients of the first three terms in the expansion of (x- a x^2 )¹², x 0 is 559. Find the value of ' a ' if ' a ' belongs to positive integers

Step-by-step solution

The expansion of (x - a x^2 )¹² is given by the binomial theorem as _ r=0 ¹² ¹²C_ r (x)^ 12-r (- a x^2 )^ r . The first three terms correspond to r=0, 1, 2 . For r=0 : Term is ¹²C₀ (x)¹² (-a)⁰ (x)⁻⁰ = 1 x¹² . The coefficient is 1 . For r=1 : Term is ¹²C₁ (x)¹¹ (-a)¹ (x)⁻² = 12 (-a) x⁹ = -12a x⁹ . The coefficient is -12a . For r=2 : Term is ¹²C₂ (x)¹⁰ (-a)² (x)⁻⁴ = 66 a² x⁶ = 66a² x⁶ . The coefficient is 66a² . The sum of these coefficients is given as 559 . 1 - 12a + 66a² = 559 66a² - 12a - 558 = 0 Dividing by 6 :