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)`
`=...
(The entire section contains 194 words.)
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`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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