Mathematical induction - divisibility - odd/even numbers (1 Viewer)

[sasquatch]

Im having troubles with the step "n = k + 2" when proving results by mathematical induction, for all odd/even n. E.g.

For even n, proven n³ + 2n is divisible by 12. Heres an example of my working.

1. If n = 2,

expression = 2<sup>3 + 2(2)
= 12, which is divisible by 12,

hence true for n = 2.

2. Assuming true for n = k, where k is an even integer.

k3 + 2k = 12m, where m is an integer

3. Hence prove true for n = k +2,

(k+2)3 + 2(k+2) is divisible by 12.

expression = (k2 + 4k + 4)(k+2) + 2k + 4
= k3 + 2k2 + 4k2 + 8k + 4k + 8 + 2k + 4
= (k3 + 2k) + 6k2 + 12k + 12
= 12m + 6k2 + 12k + 12
= 12(m + k + 1) + 6k2

Well yep thats all i can get up to... but i was thinking as |n| >= 2, would it be safe to say that,

k2 >= 4 for even n, and for any even number k, k2 is always even, hence 6k2 is divisible by 12.

Is that ok?</sup>

 

[Riviet]

You've got the idea. What you could do is:

12(m + k + 1) + 6k² = 12(m+k+1 + k²/2)

= 12A, where A is a positive integer, since k is even => k²/2 is an integer.

 

[sasquatch]

Oh yeah i see then... cool thanks