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2 câu trả lời 32
Để chứng minh rằng n3−n+3 chia hết cho 3 với mọi số tự nhiên n≥1, ta sẽ xem xét các giá trị của n theo modulo 3 và áp dụng định lý.
Bước 1: Xét các giá trị của n theo modulo 3
Mọi số tự nhiên n sẽ có dạng 3k, 3k+1, hoặc 3k+2 (với k là số nguyên không âm).
Trường hợp 1: n \equiv 0 \pmod{3}
n = 3k
n^3 - n + 3 = (3k)^3 - 3k + 3 = 27k^3 - 3k + 3
n^3 - n + 3 \equiv 0 - 0 + 3 \equiv 3 \equiv 0 \pmod{3}
Trường hợp 2: n \equiv 1 \pmod{3}
n = 3k + 1
n^3 - n + 3 = (3k + 1)^3 - (3k + 1) + 3
= 27k^3 + 27k^2 + 9k + 1 - 3k - 1 + 3
= 27k^3 + 27k^2 + 6k + 3
n^3 - n + 3 \equiv 0 + 0 + 0 + 3 \equiv 3 \equiv 0 \pmod{3}
Trường hợp 3: n \equiv 2 \pmod{3}
n = 3k + 2
n^3 - n + 3 = (3k + 2)^3 - (3k + 2) + 3
= 27k^3 + 54k^2 + 36k + 8 - 3k - 2 + 3
= 27k^3 + 54k^2 + 33k + 9
n^3 - n + 3 \equiv 0 + 0 + 0 + 9 \equiv 9 \equiv 0 \pmod{3}
Kết luận
Dựa vào các trường hợp trên, ta thấy rằng:
n^3 - n + 3 \equiv 0 \pmod{3}
Vậy n^3 - n + 3 chia hết cho 3 với mọi số tự nhiên n \geq 1 .
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