Use mathematical induction to prove that n(n2+5) is divisible by 6 for n∈N

fortdefruitI 2020-11-09 Answered

Use mathematical induction to prove that n(n2+5) is divisible by 6 for nN

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d2saint0
Answered 2020-11-10 Author has 89 answers

Let P(n) denotes that n(nn+5) is divisible by 6, for each natural number n.
P(1):1(12+5)=6, which is divisible by 6. Therefore the result is true for n=1.
Now, assume that P(n) is true for n=k. That is P(k):k(k2+5) is ÷i is  by 6.k(k2+5)=6m for some m∈N. We have to prove that P(k+1) is also true.
P(k+1):(k+1)[(k+1)2+5]

=(k+1)[k2+2k+1+5]

=(k+1)[k2+2k+6]

=k3+2k2+6k+k22k+6

=k3+3k2+8k+6

=k3+5k+3k2+3k+6

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

=(6m)+3(k2+k+2)
Since k2+k+2 is divisible by 22 for every even and odd k. Therefore 3(k2+k+2) is divisible by 66 and hence (6m)+3(k2+k+2) is divisible by 6. Therefore, P(k+1) is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.

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