# Strange Induction (1 Viewer)

#### fan96

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

#### HoldingOn

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

#### Drongoski

##### Well-Known Member
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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