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  • n(n2 + 5) is divisible by 6, for each natural number n.

n(n^2 + 5) is divisible by 6, for each natural number n.

Answers (1)

P(n)= n(n^2 + 5) is divisible by 6, for each natural number n.

Now, we’ll substitute different values for n,

P(1) = 1(1^2 + 5) = 6, is divisible by 6

P(2) = 2(2^{}2 + 5) = 18, is divisible by 6

Let us consider,

P(k) = k(k^2 + 5)be divisible by 6

Thus, k(k^2 + 5)=6x

We also get that,

P(k+1) = (k+1)[(k+1)^{2} + 5]

          = (k+1)(k^2 + 2k + 6)

            = k(k^2 + 5 ) + k(2k+1) + k^2+2k+ 6

           = 6x + 3k^2 + 3k + 6  \because (k^2 + k ) \text{ is even }

         = 6x + 3\times 2y + 6, is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= n(n^2 + 5) is divisible by 6.

Posted by

infoexpert21

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