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how do i prove that (3n+1)^2 - (3n - 1)^2 is a multiple of 6

how do i prove that (3n+1)^2 - (3n - 1)^2 is a multiple of 6?

Im really stuck, i have expanded out the brackets but i don't know what to do next...


Thanks in advance :smile:
After expanding and sorting signs out... 9n^2+6n+1-9n^2+6n-1=12n=2(6n) which proves for any input n, the result will be a multiple of 6
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Original post by Whyhuyrah
After expanding and sorting signs out... 9n^2+6n+1-9n^2+6n-1=12n=2(6n) which proves for any input n, the result will be a multiple of 6


where does 2(6n) come from?
Collect the like terms together, and remember that it is (3n+1)^2 - (3n - 1)^2, so all of the terms from (3n - 1)^2 are affected by the subtraction sign.
Original post by Whyhuyrah
After expanding and sorting signs out... 9n^2+6n+1-9n^2+6n-1=12n=2(6n) which proves for any input n, the result will be a multiple of 6


You could even say 6(2n)
I factored out the 2, and yes, you can also factor out the 6, however, I interpreted the proof to follow (3n+1)^2-(3n-1)^2=k(6n) (Where k is a constant to be determined) so it showed the n is multiplied by 6 first, it wouldn't really matter either way as long as you concluded with a valid statement

You could also tackle it as the difference of two squares, a lot simpler
(3n+1)^2-(3n-1)^2=(3n+1+3n-1)(3n+1-3n+1)=(6n)2
(edited 8 years ago)

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