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p34_19_correction
𝐴 = 6𝑛 + 5 = 2 × 3𝑛 + 4 + 1
= 2 × 3𝑛 + 2 × 2 + 1
= 2(3𝑛 + 2) + 1 = 2 × 𝑒𝑛𝑡𝑖𝑒𝑟 + 1 donc c’est un impair.
𝐵 = (5𝑛 + 2) + (3𝑛 − 1) − (4 − 4𝑛)
= 5𝑛 + 2 + 3𝑛 − 1 − 4 + 4𝑛
= 12𝑛 − 3
= 2 × 2𝑛 − 4 + 1
= 2 × 2𝑛 − 2 × 2 + 1
= 2(2𝑛 − 2) + 1
= 2 × 𝑒𝑛𝑡𝑖𝑒𝑟 + 1
𝐶 = (4𝑛 + 1)(4𝑛 − 1) − 4
= 16𝑛2 − 1 − 4
= 16𝑛2 − 5
= 16𝑛2 − 6 + 1
= 2 × 8𝑛2 − 2 × 3 + 1
= 2 × (8𝑛2 − 3) + 1
= 2 × 𝑒𝑛𝑡𝑖𝑒𝑟 + 1