Zadanie 1f, Paper 3, maj 2022, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2022, TZ1, Paper 3 (zadania wzorowane), pytanie 1(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A counter-building scheme starts with one isolated central token. For a fixed positive integer r, layer j adds 2rj tokens, for j=1,2,… . After n−1 layers the count is , with n a positive integer; an empty sum is 0. Equivalently Cr(n) = 1 + r(n2 − n). For r=1 the first counts are 1,3,7,13,21.
Polecenie
A weighted version of the counter scheme has , for positive integers r,n. Prove by mathematical induction that Dr(n) = n(n + 1)[2r(2n + 1) + 3]/6.
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At n=1, the sum is 2r+1, and the formula gives 2(6r+3)/6=2r+1. Assume the formula holds at n=k. Add the next term: Dr(k + 1) = k(k + 1)[2r(2k + 1) + 3]/6 + (k + 1)[2r(k + 1) + 1]. Factor (k+1)/6: the remaining bracket is k(4rk + 2r + 3) + 12r(k + 1) + 6 = 4rk2 + (14r + 3)k + 12r + 6. This factors as (k + 2)(4rk + 6r + 3) = (k + 2)[2r(2(k + 1) + 1) + 3]. Therefore D_r(k+1) has precisely the claimed form at k+1. Together with the base case, this proves the identity for all positive n.
Schemat punktowania
R1:both sides at n=1. M1:explicit induction hypothesis at k. M1:add the correct next term. A1:correct substitution and common denominator. M1:factor or rearrange toward the k+1 expression. A1:correct expanded bracket 4rk²+(14r+3)k+12r+6. A1:correct factorization yielding the k+1 formula. R1:valid induction conclusion, dependent on a valid base/hypothesis/step. A direct sum without induction does not meet the prescribed method.
Komentarz. Paper 3: graphic display calculator and AA HL formula booklet allowed. Proofs require reasoning, not isolated numerical checks. Credit valid alternative methods and follow-through unless inconsistent with a supplied result.