Answer :

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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