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 Question from John, a student: I am very lost on this one: Prove: For all n in Natural Numbers ( n > 1 ), n^3/3 + n^5/5 + 7n/15 is an integer i. Let n=1, then 1/3 + 1/5 + 7/15 = (5 + 3 + 7 )/15 = 15/15 = 1 . True ii. assume n^3/3 + n^5/5 + 7n/15 is an integer show this is also true for ( n+1). I can't figure out how to set this up so that I can use the induction hypothesis to make the assumption useful in my proof. Thank You,

Hi John.

Begin with (n+1)3/3 + (n+1)5/5 + 7(n+1)/15

= (1/3)( n3 + 3n2 + 3n + 1) + (1/5)(n5 + 5n4 + 10n3 + 10n2 + 5n + 1) + 7n/15 + 7/15
= [ n3/3 + n5/5 + 7n/15 ] + (1/3)( 3n2 + 3n + 1) + (1/5)( 5n4 + 10n3 + 10n2 + 5n + 1) + 7/15
= [an integer] + (1/3)( 3n2 + 3n + 1) + (1/5)( 5n4 + 10n3 + 10n2 + 5n + 1) + 7/15

and you have to demonstrate that the rest is an integer. Can you finish the algebra?

Cheers,
Stephen La Rocque.

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