# Prove the followi

Let P(n): n(n + 1) (n + 5) is multiple by 3 for all nN

Let P(n) is true for n=1

P(1): 1(1 + 1) (1 + 5)

= 2 × 6

= 12

Since, it is multiple of 3

So, P(n) is true for n = 1

Now, Let P(n) is true for n = k

P(k): k(k + 1) (k + 5)

= k(k + 1) (k + 5) is a multiple of 3

Then, k(k + 1) (k + 5) = 3λ - - - - - (1)

We have to show,

= (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

= (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

Now,

= (k + 1)[(k + 1) + 1][(k + 1) + 5]

= (k + 1)(k + 2)[(k + 1) + 5]

= [k(k + 1) + 2(k + 1)][(k + 5) + 1]

= k(k + 1)(k + 5) + k(k + 1) + 2(k + 1)(k + 5) + 2(k + 1)

= 3λ + k2 + k + 2(k2 + 6k + 5) + 2k + 2

= 3λ + k2 + k + 2k2 + 12k + 10 + 2k + 2

= 3λ + 3k2 + 15k + 12

= 3(λ + k2 + 5k + 4)

= 3μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all nN

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