Strange Induction (1 Viewer)

[HoldingOn]

http://prntscr.com/khe3aj

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

 

[fan96]

Have you tried expanding and rearranging?

 

[HoldingOn]

Have you tried expanding and rearranging?

Yes

 

[Drongoski]

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.