Q. 2

# Prove that if n i

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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