Verified answer

yes

(3n+1)² - (3n-1)² =9n^2+6n +1 -9n^2 +6n -1 =12n =4*3n

Multiply out both binomials:

(9n^2 + 6n + 1) - (9n^2 - 6n + 1) = 9n^2 + 6n + 1 - 9n^2 + 6n -1 = 12n.

12n = 4*(3n) so the sum of the two binomials is always going to be 4*(3n), which is a multiple of 4.

let's view (3n+1)^2 as a^2 and (3n-1)^2 and b^2, so that we can implement a^2 - b^2 formula to it.

(3n+1)² - (3n-1)² = (3n+1+3n-1)(3n+1-3n+1)

= (6n)*2

= 12n

= 4*3n Hence proved.. :)

Notice that (3n+1)^2 - (3n-1)^2 = (9n^2+6n+1)-(9n^2 - 6n + 1) = 12n = 4*(3n).

Hence, it is a multiple of 4.

[ ( 3n + 1) - ( 3n - 1 ) ] [ ( 3n + 1) + ( 3n - 1 ) ]

2 x 6n

12 n which is a multiple of 4

(3 n + 1)^2 - (3 n - 1)^2

=

12 n

(4 x 3) n

fail