For any positive integer n, prove that n3 – n is divisible by 6. no spam plz ❌❌.
Answers
Step-by-step explanation:
The condition for any number to be divisible by 6 is that the number must be individually divisible by 3 and 2.
Check whether n
3
−n is divisible by 3.
n
3
−n=n(n+1)(n−1)
When a number is divided by 3 then by the remainder theorem, the remainder obtained is either 0 or 1 or 2.
n=3p or n=3p+1 or n=3p+2, where p is some integer.
If n=3p, then the number is divisible by 3.
If n=3p+1, then n−1=3p+1−1=3p. The number is divisible by 3.
If n=3p+2, then n+1=3p+2+1=3(p+1). The number is divisible by 3.
So, any number in the form of n
3
−n=n(n+1)(n−1) is divisible by 3.
Check whether n
3
−n is divisible by 2.
When a number is divided by 2, the remainder obtained is either 0 or 1 by the remainder theorem.
n=2p or n=2p+1, where p is some integer.
If n=2p, then the number is divisible by 2.
If n=2p+1 then n−1=2p+1−1=2p. The number is divisible by 2.
So, any number in the form of n
3
−n=n(n+1)(n−1) is divisible by 2.
Since, the given number n
3
−n=n(n+1)(n−1) is divisible by both 3 and 2. Therefore, according to the divisibility rule of 6, the given number is divisible by 6.
Hence, n
3
−n=n(n+1)(n−1) is divisible by 6.
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.
Step-by-step explanation:
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