for any positive integer n prove that n cube minus n is divided by 6
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Concept used:
MATHEMATICAL INDUCTION
Step-by-step explanation:
⇒ Let n = 1.
- n³ - n
- 1³ - 1
- 1 - 1
- 0
- As any number divides 0 completely (with remainder 0), so does 6.
⇒ Let n = k.
- Assume that k³ - k is divisible by 6.
⇒ Let n = k + 1.
- n³ - n
- (k + 1)³ - (k + 1)
- (k³ + 3k² + 3k + 1) - (k + 1)
- k³ + 3k² + 3k + 1 - k - 1
- k³ - k + 3k² + 3k
- k³ - k + 3k(k + 1)
- Here, 3k(k + 1) is 3 multiplied to the product of two consecutive natural numbers. As product of two consecutive integers is even, then so is k(k + 1), so that 3k(k + 1) is multiple of 6. In k³ - k + 3k(k + 1), we assumed earlier that k³ - k is a multiple of 6. To this, 3k(k + 1), which is a multiple of 6, is added.
⇒ Hence proved!!!
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