A-LevelFurther MathematicsProof by induction9231/12 · Oct/Nov 2014 · Q37 Marks

It is given that u_r = r × r! for r = 1, 2, 3, ... . Let S_n = u_1 + u_2 + u_3 + ... + u_n. Write down the values of 2! - S_1, 3! - S_2, 4! – S_3, 5! – S_4. Conjecture a formula for S_n. Prove, by mathematical induction, a formula for S_n, for all positive integers n.

mainIt is given that u_r = r × r! for r = 1, 2, 3, ... . Let S_n = u_1 + u_2 + u_3 + ... + u_n. Write down the values of 2! - S_1, 3! - S_2, 4! – S_3, 5! – S_4. Conjecture a formula for S_n. Prove, by mathematical induction, a formula for S_n, for all positive integers n.[7]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

[B2,1,0] 2!-S₁ = 1, 3!-S₂ = 1, 4!-S₃ = 1, 5!-S₄ = 1 (Two correct B1, all four correct B2)
[B1] S_n = (n+1)! - 1
[B1] For n=1, S_1 = 1 × 1! = 1. Formula gives (1+1)! - 1 = 2! - 1 = 1. So H_1 is true.
[B1] Assume H_k: S_k = (k+1)! - 1
[M1] S_{k+1} = S_k + u_{k+1} = (k+1)! - 1 + (k + 1) × (k + 1)! = (k+1)!(1 + k + 1) - 1 = (k+1)!(k+2) - 1 = (k+2)! - 1
[A1] Hence H_k ⇒ H_{k+1}. So result holds for all positive integers (by PMI).

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