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in Mathematical Induction by (32.1k points)
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Prove the following by the principle of mathematical induction: 

n(n + 1) (n + 5) is a multiple of 3 for allnϵ N. 

Show that: P(n): n(n + 1) (n + 5) is multiple by 3 for all n∈N

1 Answer

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

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 + 3k 2 + 15k + 12 

= 3(λ + k 2 + 5k + 4) 

= 3μ 

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

Hence, P(n) is true for all n∈N

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