3i will delete my post.
Prove that n/3+n2/2+n3/6 is an integer for all integers n!
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5 Answers
1given equation is = (2n+3n2+n3)/6
(n)(n+1)(n+2)/6
now the numerator is a product of 3 consecutive integers
so it will be a multiple of 2 and 3 so a multiple of 6
hence the given equation is an integer for all n
341\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6} will be an integer if \frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6} - n = \left(\frac{n}{3}-\frac{n}{3}\right ) + \left(\frac{n^2}{2} - \frac{n}{2} \right) + \left(\frac{n^3}{6}- \frac{n}{6} \right) is an integer
Since n(n-1) is divisible by 2 and n3-n = (n-1)(n)(n+1) is divisible by 6, the result is true.
This is a more general method I feel, because I have seen such qns with higher powers where factorisation will not be an easy option