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  • 9. n3 -n is divisible by 6, for each natural number n.

n^3 -n is divisible by 6, for each natural number n. n \geq2

Answers (1)

P(n)= n^3 -n is divisible by 6

Now, we’ll substitute different values for n,

P(2) = 2^3 -2 = 6, is divisible by 6

Now, let us consider,

P(k) = k^3 -k be divisible by 6

Thus, k^3 -k = 6x

We also get that,

P(k+1) =(k+1)^3 -(k+1)

= (k+1)(k^2 + 2k + 1 -1)

= k^3 + 3k^2 + 2k

= 6x + 3k(k+1)  

= 6x + 3 \times 2y,   \because k(k+1) \text{ is even}

= 6x + 3 \times 2y 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^3 -n is divisible by 6.
 

Posted by

infoexpert21

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