Prove that (3n+ 1)^2– (3n-1)^2is a multiple of 4, for all positive integer
values of n.

Step-by-step explanation:

Using a² - b² = (a + b)(a - b), we have

(3n + 1 + 3n - 1)(3n + 1 - 3n + 1)

= 6n * 2 = 12n.

Since 12n can be expressed as 4(3n) and 3n is an integer, we have proven it must be a multiple of 4.

the answer on e2020 was d.