Anonymous

P = (3k+1)^2 - (3k-1)^2 = (3k)^2 + 2(3k) + 1 -[(3k)^2 -2(3k) +1] = 0 = 0*6. Clearly, P is a multiple of 6.

Anonymous

[ (3k + 1) - (3k - 1) ] [ (3k + 1) + (3k - 1) ] (3k + 1)² - (3k - 1)² = 9k² + 6k + 1 - [ 9k² - 6k + 1 ] = 12 k 12k is a multiple of 6

Anonymous

I know how to do this. Many years ago. They are asking you to figure it out otherwise you will not be allowed to pump gas.

Anonymous

(3k+1)^2-(3k-1)^2 = (3k+1+3k-1)(3k+1-3k+1) = 6k(2) = a multiple of 6

Anonymous

k is a positive integer (3k + 1)^2 - (3k - 1)^2 = 6k(2) = 12k (a multiple of 6)

Anonymous

(3k + 1)^2 - (3k - 1)^2 = (3k + 1 + 3k - 1)(3k + 1 - 3k + 1) = 6k * 2, or 12k. Is (3k + 1)^2 - (3k - 1)^2 a multiple of 6 if k is a positive integer? YES.

Anonymous

(3k + 1)^2 - (3k - 1)^2 = = (9k^2 + 6k + 1) - (9k^2 - 6k + 1) = = 9k^2 + 6k + 1 - 9k^2 + 6k - 1 = = 12k <=== *Note: 12 is a multiple of 6 and k is a (positive) integer, so 12 times k has to be a multiple of 6 too.