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Strange Induction (1 Viewer)
- Thread starter HoldingOn
- Start date
HoldingOn
Active Member
YesHave you tried expandingand rearranging?
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.
= (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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