Prove that (3n+1)²-(3n-1)² is a multiple of 4 taking into account that n is a positive integer value

  1. 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
NK
Answered by Nalin K. Maths tutor

6471 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do I find the roots and and coordinates of the vertex of the graph y = 2x^2 + 4x - 8 ?


Simplify 8x-3+6x


Pythagoras' Theorem


Use the quadratic formula to find the solutions to x^2+2x-35=0


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences