Prove Even Integer for Natural Number n: Induction

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The discussion focuses on proving that the expression \(\frac{n^{3}+5n}{3}\) is an even integer for all natural numbers \(n\) using mathematical induction. The proof begins by substituting \(n+1\) into the expression, leading to the equation \((n+1)^3 + 5(n+1) = n^3 + 5n + 3n(n+1) + 6\). This transformation clearly shows that the result remains divisible by 3, confirming the evenness of the expression. Participants confirm their understanding of the induction process and express gratitude for the clarification provided.

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Use mathematical induction, to prove that \frac{n^{3}+5n}{3}


is an even integer for each natural number n.

I am fimilar with proof by induction but in most of the question that I have done have a
LHS = RHS which seems to simplifiy things a little bit.
Any help would be appreciated
Cheers
 
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asset101 said:
Use mathematical induction, to prove that \frac{n^{3}+5n}{3}
is an even integer for each natural number n.

I am fimilar with proof by induction...

Put n+1 in place of n.

(n+1)^3 + 5(n+1) = n^3+3n^2+3n+1+5n+5 = (n^3+5n) + 3n(n+1) + 6.

Now divide each term by 3 and see what kind of number you get.

Since you are familiar with induction, this should be enough.
 
Got it thanks mate
 

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