# Thread: Strange Induction

1. ## Strange Induction

http://prntscr.com/khe3aj

Thanks. It looks easy but... maybe I am missing something.

2. ## Re: Strange Induction

Have you tried expanding $(k+1)^3 + 5(k+1)$ and rearranging?

3. ## Re: Strange Induction

Originally Posted by fan96
Have you tried expanding $(k+1)^3 + 5(k+1)$ and rearranging?
Yes

4. ## Re: Strange Induction

this expression = k3 + 3k2 + 3k ++ 1 + 5k + 5

= (k3 + 5k) + 3k(k+1) + 6

= 6M + 6 + 3k(k+1) (M an integer)

But k(k+1) is always even, since if k is even, then it is; if k is odd, then k+1 is even. .: k(k+1) = 2N (N integer)

.: the original expression becomes 6(M + 1 + N) = 6 x integer.

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