Question

For any positive integer n, prove that n3 ‒ n is divisible by 6.the attachment is my drawimg ​

Answer 1

Answer:

Step-by-step explanation:

Let :-

a = n³ - n

= n(n² - 1)

= n(n - 1)(n + 1)

= (n - 1)n (n + 1)

1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 2

2) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3

From (1) and (2)

a must be divisible by 2 × 3 = 6

Hence n³ - n is divisible by 6 for any positive integer n

Answer 2

Answer:

n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.

Now, we have to show that the product of three consecutive positive integers is divisible by 6.

We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.

Now three consecutive positive integers are n, n + 1, n + 2.

Case I. If n = 3q.

n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)

But we know that the product of two consecutive integers is an even integer.

∴ (3q + 1) (3q + 2) is an even integer, say 2r.

⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.

Case II. If n = 3n + 1.

∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)

= (even number say 2r) (3) (q + 1)

= 6r (q + 1),

which is divisible by 6.

Case III. If n = 3q + 2.

∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)

= multiple of 6 for every q

= 6r (say),

which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6.