Question

For any positive integer n, prove that
n^2-n is divisible by 6

Answer 1

Let n be any positive integer

now applying division lemma on a and 6 we have

n = 6q + r[r = 0,1,2,3,4,5]

if n = 6q,

n²-n = (6q) ² - 6 q

= 36q² - 6q= 6(6q²-q) which is divisible by 6

similarly

n²-n is divisible by 1,2,3,4,5

so by mathematical induction

n²-n is divisible by 6

hope it helps mark as brainliest

Answer 2

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