Zadanie 1f, Paper 3, maj 2023, TZ1, Matematyka AA IB HL

Base n=1: A_1=1/2=1!/2. Assume A_k=k!/2. For k+1, integrate $A_{k+1}=\int_0^\infty x^{2k+2}(xe^{-x^2})\,dx$ by parts. With u=x^(2k+2) and v=−e^(−x²)/2, this is $[-x^{2k+2}e^{-x^2}/2]_0^\infty+(k+1)\int_0^\infty x^{2k+1}e^{-x^2}\,dx$. The boundary term is zero at both ends by the supplied limit and k≥1. Hence A_{k+1}=(k+1)A_k=(k+1)k!/2=(k+1)!/2. The base case and induction step prove the formula for all positive integers n.

M1:consider the base case. A1:verify both sides equal 1/2. M1:state the induction hypothesis. M1:attempt integration by parts for A_{k+1}, showing a boundary expression and an integral. A1:correct boundary term and recurrence coefficientk+1. A1:use the hypothesis to obtain (k+1)k!/2. A1:simplify to (k+1)!/2. R1:valid induction conclusion, requiring at least four preceding marks. Starting at n=0 with A_0=1/2 is also valid. Merely writing let n=k does not earn the induction-hypothesis M1; subsequent marks remain independently available. Condone omission of the explicit zero boundary term.