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GeoMike
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The following problem is given in my algebra book:
For all positive integers n, n(n+1)(n+2) is divisible by 6. Prove using mathematical induction.
First:
k=1, 1(1+1)(1+2) = 1(2)(3) = 6, which is obviously divisible by 6
Next:
k(k+1)(k+2)
And then:
(k+1)(k+2)(k+3)
From this I distributed the (k+3) to get:
k(k+1)(k+2) + 3(k+1)(k+2)
The first term is the assumed case (k, divisible by 6)
But 3(k+1)(k+2) doesn't appear to be a clear case of being divisible by 6.
So... I multiplied it out:
[tex]3(k+1)(k+2) = 3(k^2+3k+2) = 3k^2+9k+6[/tex]
From here I took [tex]3k^2+9k+6[/tex] and restarted the process, so:
For k=1, [tex]3(1)^2+9(1)+6 = 3+9+6 = 18[/tex], which is divisible by 6
The k case:
[tex]3k^2+9k+6[/tex]
And the k+1 case:
[tex]3(k+1)^2+9(k+1)+6 = 3(k^2+2k+1)+9k+9+6 = 3k^2+6k+3+9k+9+6[/tex]
Collecting terms:
[tex](3k^2+9k+6) + (6k+12)[/tex], The first part being the assumed case (k), and the second part is obviously divisible by 6
So I worked it and was able to prove it (I think, unless I did something wrong). What I want to know is: is there an easier way to do this (using just the basic mathematical induction that the book wants)? It seems like I took more steps than necessary/over-complicated the proof. Could I have made it more concise? Is it ok as is?
Thanks!
GM
For all positive integers n, n(n+1)(n+2) is divisible by 6. Prove using mathematical induction.
First:
k=1, 1(1+1)(1+2) = 1(2)(3) = 6, which is obviously divisible by 6
Next:
k(k+1)(k+2)
And then:
(k+1)(k+2)(k+3)
From this I distributed the (k+3) to get:
k(k+1)(k+2) + 3(k+1)(k+2)
The first term is the assumed case (k, divisible by 6)
But 3(k+1)(k+2) doesn't appear to be a clear case of being divisible by 6.
So... I multiplied it out:
[tex]3(k+1)(k+2) = 3(k^2+3k+2) = 3k^2+9k+6[/tex]
From here I took [tex]3k^2+9k+6[/tex] and restarted the process, so:
For k=1, [tex]3(1)^2+9(1)+6 = 3+9+6 = 18[/tex], which is divisible by 6
The k case:
[tex]3k^2+9k+6[/tex]
And the k+1 case:
[tex]3(k+1)^2+9(k+1)+6 = 3(k^2+2k+1)+9k+9+6 = 3k^2+6k+3+9k+9+6[/tex]
Collecting terms:
[tex](3k^2+9k+6) + (6k+12)[/tex], The first part being the assumed case (k), and the second part is obviously divisible by 6
So I worked it and was able to prove it (I think, unless I did something wrong). What I want to know is: is there an easier way to do this (using just the basic mathematical induction that the book wants)? It seems like I took more steps than necessary/over-complicated the proof. Could I have made it more concise? Is it ok as is?
Thanks!
GM
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