Zadanie 1b, Paper 3, maj 2022, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2022, TZ1, Paper 3 (zadania wzorowane), pytanie 1(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
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
(i) Show that C1(n) + C1(n + 1) = 2n2 + 2. (ii) Describe this result in words as a rearrangement of tokens into two equal square arrays and isolated tokens. (iii) For n=3 sketch such a rearrangement, distinguishing tokens from the two counts.
Korepetytor
Nie wiesz, jak ruszyć? Korepetytor zna to zadanie i poprowadzi Cię krok po kroku, bez gotowca.
3 rozmowy · bez karty · Zaloguj się
Sprawdź się
Jak poszło?
Pokaż rozwiązanieUkryj rozwiązanie
Odpowiedź z klucza
(i) n(n−1)+1+(n+1)n+1=2n²+2. (ii) Two consecutive counts can be rearranged into two n-by-n square arrays and two isolated tokens. (iii) The counts are 7 and 13. Draw two 3×3 arrays and two isolated tokens: mark six cells in the first array plus one isolated token with one symbol; mark the remaining three cells there, all nine cells of the second array and the other isolated token with another symbol. The totals are 7 and 13, and 20 overall.
Schemat punktowania
(i) M1:add the two explicit formulas. A1:correct simplification. (ii) A1:two equal square arrays plus two single tokens. (iii) A1:diagram with 7+13 distinct tokens arranged as two 3×3 arrays plus two singles; equivalent arrangements accepted.
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.