# proof by induction that n(n² + 5) can be divided by 6?

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`f(n) = n(n^2+5) `

We have to prove that f(n) is divisible by 6.

We can use mathematical induction here.

`f(n) = n(n^2+5)`

For n = 1

`f(1) = (1)(1+5) = 6 = 6xx1 `

So when n = 1 f(n) is divisible by 6.

Let us assume that when n = p where p>1 the result is true.

`f(p) = p(p^2+5) = 6xxK` where K is a positive integer.

`f(p+1) `

`= (p+1)((p+1)^2+5)`

`=...

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## Related Questions

aruv | Student

`P(n): n(n^2+1)` is divisible by 6 .

P(n): stament that says n(n^2+1) is divisible by 6

when n=1

`P(1): 1(1^2+5)=1(1+5)=1.6=6`

6 is divisible by 6. so statement P(1) is true.

Let  statement is true for n=k i.e.

`P(k): k(k^2+5)`  is divisible by 6 .

To prove P(n) is true when n=k+1

`P(k+1): (k+1)((k+1)^2+5)`

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

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

`=k(k^2+5)+(3k^2+3k+6)`

`=k(k^2+5)+3(k^2+k+2)`

`k^2+k+2` is always an even number say 2M ( M is an integer).

`k^2+k+2=2M`

Thus

`P(k+1): (k+1)((k+1)^2+5)=k(k^2+5)+3.2M`

`k(k^2+5)`  is divisible by 6 since P(k) is true.

6M is divisible by 6.

Thus

`k(k^2+5)+6M`  is divisible by 6

Thus P(k+1) is true when P(k) is true.

Thus P(n) is true for all n.

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