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n(n^2 + 5) is divisible by 6 for all integers n >= 1.
i am kind of stuck here
Originally Posted by neelpatel89
Where EXACTLY are you stuck? Have you already checked for n = 1? Then, what is your inductive asusmption (hypothesis)?
It's possible to do it without induction, too:
(1) if n is an odd multiple of 3 then n^2 is odd and thus n^2 + 5 is even ==> the product n(n^2 + 5) is divisible both by 2 and 3 and thus by 6
(2) if n is an even multiple of 3 then it is a multiple of 6...
(3) if n is NOT a multiple of n the n^2 + 5 is ALWAYS a multiple of 3 (right n = 1 or 2 + 3k,. with k an integer, and square this and add to 5): if n is odd then n^2 + 5 is even, and if n is even good: in any case, the product n(n^2 + 5) is divisible both by 2 and by 3.
Tonio
It's obvioulsy true for n = 1.
Assume true for n = m.
Show true for n = m implies true for n = m+1:
The first term is divisible by 6 (from the inductive hypothesis). Your job is to prove that the second term, that is,is divisible by 6. I suggest using induction ....
Or simpler, just divide in two cases: odd m and even m. In both cases the expression within the parentheses in even and we're doneIt's obvioulsy true for n = 1.
Assume true for n = m.
Show true for n = m implies true for n = m+1:
The first term is divisible by 6 (from the inductive hypothesis). Your job is to prove that the second term, that is,
Tonio
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