Hey,
I need to proove by Induction that k^3+3k^2+2k devided by 6 without residue.
thanks
omri
you can actually check that $\displaystyle k^3+3k^2+2k = k(k+1)(k+2)$ and for any natural number k, one of them is even, and one is divisible by 3. thus, it must be divisible by 6.
by induction:
for k=1: 1 + 3 + 2 = 6
suppose true for n < k+1
for n=k+1: $\displaystyle (k+1)^3 + 3(k+1)^2 + 2(k+1) =k^3+3k^2+3k+1+3k^2+6k+3+2k+2$
$\displaystyle =k^3+6k^2+11k+6 = k^3 + 3k^2 + 2k + 3k^2 + 9k + 6$
$\displaystyle =k(k^2+3k+2) + 3(k^2+3k+2) = (k+3)(k^2+3k+2)=(k+3)(k+2)(k+1)$
which is similar to the equation above..