Question

Let A = n3 + 6n2 + 8n. Prove that A is divisible by 3, Vn EN​

Answer 1

Answer:

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Answer 2

Given :  A = n3 + 6n2 + 8n.

To Find : Prove that A is divisible by 3

Solution:

A = n³  + 6n²  + 8n

= n(n² + 6n + 8)

=  n (n + 2)(n + 4)

without loosing generality

n = 3k , 3k +1 , 3k + 2    k ∈ Z

n = 3k

=> A = 3k (3k + 2)(3k + 4)

Hence divisible by 3

n = 3k + 1

=> A = (3k + 1)((3k + 1 + 2)(3k + 1 + 4)

= (3k + 1)(3k + 3)(3k + 5)

= 3(3k + 1)(k + 1)(3k + 5)

Hence divisible by 3

n = 3k + 2

=> A = (3k + 2)(3k + 2 + 2)(3k +2  + 4)

= (3k + 2)(3k + 4)(3k + 6)

= 3(3k + 2)(3k + 4)(k + 2)

Hence divisible by 3

so A = n³  + 6n²  + 8n  is divisible by 3

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