[14] Question Scheme Marks Number 8. (a) If =1, and , n B1 (so true for = 1. Assume true for = n n k) So M1 = = A1 = which implies is true for dA1 As result is true for this implies true for all positive integers and so dM1A1cso result is true by induction (6) (b) B1 (so true for . Assume true for = n k) M1, which implies is true for = + 1 A1 n k As result is true for = 1 this implies true for all positive integers and M1A1cso n so result is true by induction (5) [11] Question Scheme Marks Number 9. (a) so M1 Gradient when = 4 is and gradient of normal is M1 A1 x So equation of normal is (or ) M1 A1 (5) (b) is at point (9,0) B1 S is at (22,0), found by substituting = 0 into their part (a) B1ft N y
[14] Question Number Scheme Marks 8. (a) If n =1, and , B1 (so true for n = 1. Assume true for n = k) So M1 = = A1 = which implies is true for dA1 As result is true for this implies true for all positive integers and so result is true by induction dM1A1cso (6) (b) B1 (so true for . Assume true for n = k) M1, which implies is true for n = k + 1 A1 As result is true for n = 1 this implies true for all positive integers and so result is true by induction M1A1cso (5) [11] Question Number Scheme Marks 9. (a) so M1 Gradient when x = 4 is and gradient of normal is M1 A1 So equation of normal is (or ) M1 A1 (5) (b) S is at point (9,0) B1 N is at (22,0), found by substituting y = 0 into their part (a) B1ft
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