Prove by the principle of mathematical induction:n(n + 1) (n + 5) is a

1.	Prove by the principle of mathematical induction:n(n + 1) (n + 5) is a multiple of 3 for all n ϵ N.
Answer» Suppose P (n): n(n + 1) (n + 5) is a multiple of 3 Now let us check for n = 1, P(1): 1(1 + 1) (1 + 5) : 2 × 6 : 12 P (n) is true for n = 1. Where, P (n) is a multiple of 3 Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true. P (k): k(k + 1) (k + 5) is a multiple of 3 : k(k + 1) (k + 5) = 3λ … (i) Now we have to prove, (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3 (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ Therefore, = (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 P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction:n(n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

Answer»

Suppose P (n): n(n + 1) (n + 5) is a multiple of 3

Now let us check for n = 1,

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

: 2 × 6

: 12

P (n) is true for n = 1. Where, P (n) is a multiple of 3

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): k(k + 1) (k + 5) is a multiple of 3

: k(k + 1) (k + 5) = 3λ … (i)

Now we have to prove,

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

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

Therefore,

= (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

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

Thus, P (n) is true for all n ∈ N.