Prove that (3n+1)²-(3n-1)² is a multiple of 4 taking into account that n is a positive integer value
- Square the brackets (3n+1)²= (3n+1)(3n+1) = 9n²+3n+3n+1 = 9n²+6n+1 (3n-1)²= (3n-1)(3n-1) = 9n²-3n-3n+1 = 9n²-6n+12. Write out the full equation (9n²+6n+1) - (9n²-6n+1) = 9n²+6n+1-9n²+6n-1 = 12n3. Explain your reasoning 12n is divisible by 4 as (12n÷4) equals 3n therefore (3n+1)²-(3n-1)² is a multiple of 4 as 4 goes into 12 a total of 3 times and 3 is an integer
