Q- Prove that some of square of two odd positive
Q- Show that for any positive integer n, n³-n is divisible by 6
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Answered by
2
Answer:
Let us consider
a = n3 – n
a = n (n2 – 1)
a = n (n + 1)(n – 1)
Assumtions:
1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2.
2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.
From (1) and (2) a must be divisible by 2 × 3 = 6
Thus, n³ – n is divisible by 6 for any positive integer n.
Answered by
6
Answer:
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