Q. 2

Prove that if n is a positive even integer, then 24 divides n(n + 1)(n + 2).

Answer :

Given here, n is a positive even integer.

We know that any positive even integer can be expressed as n = 2k


Therefore, n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)……eq (1)


Now, considering the following cases,


Case 1: k = 1


n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)


= 2 × 1(2 × 1 + 1)(2 × 1 + 2)


= 2 × (3) × (4)


= 24


Hence, n(n + 1)(n + 2) is divisible by 24. (From (1))


Case 2: k = 2


n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)


= 2 × 2(2 × 2 + 1)(2 × 2 + 2)


= 4 × (5) × (6)


= 24 × 5


Hence, n(n + 1)(n + 2) is divisible by 24. (from (1))


Case 3: k3


Here, k and k + 1 being even integers, one of them will always be odd and the other will be even.


So, 2k(2k + 1)(2k + 2) will always be divisible by 2.


Also as k≥ 3 is a positive integer, so for some lN,


k = 3a or k = 3a + 1 or k = 3a + 2


For k = 3a


k(k + 1)(2k + 1) = 3a(3a + 1)(2(3a) + 1)


= 3(a(3a + 1)(6a + 1)


Therefore, it is divisible by 3.


For k = 3a + 1


k(k + 1)(2k + 1) = (3a + 1)(3a + 1 + 1)(2(3a + 1) + 1)


= (3a + 1)(3a + 2)(6a + 3)


= 3(3a + 1)(3a + 2)(2a + 1)


Therefore, it is divisible by 3.


For k = 3a + 2


k(k + 1)(2k + 1) = (3a + 2)(3a + 2 + 1)(2(3a + 2) + 1)


= (3a + 2)(3a + 3)(6a + 5)


= 3(3a + 2)(a + 1)(6a + 5)


Therefore, it is divisible by 3.


We can see that in any case k(k + 1)(2k + 1) is divisible by 2 and 3.


Also, we know that 2 and 3 are mutually prime numbers.


Therefore the expression must be divisible by 2 × 3 = 6


So, k(k + 1)(2k + 1) is divisible by 24. (From (1))


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