Show that (n^2) + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5, Hence show that the sum of 3 consecutive square numbers is always 2 away from a multiple of 3.

Expanding out the Brackets: (n2)+ (n2 + n + n + 1) + (n2 + 2n + 2n +4) = (n2) + (n2+2n+1) + (n2+4n+4) =3n2 + 6n +5 Using this result: Three consecutice numbers: n, n+1, n+2 So three consecutive squares:  n2, (n+1)2, (n+2)2 Sum of those brackets = 3n2 + 6n + 5 (from first part) Simplifying 3n2 + 6n + 5 = 3n2 + 6n + 3   +2 = 3(n2 + 2n +1)  + 2 First part is always a multiple of 3, and the final result is always 2 away from that.

JC
Answered by James C. Further Mathematics tutor

2528 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

If y=(x^2)*(x-10), work out dy/dx


A straight line passes trough the points A(-4;7); B(6;-5); C(8;t). Use an algebraic method to work out the value of t.


f(x) = 2x^3+6x^2-18x+1. For which values of x is f(x) an increasing function?


How can I show that the lines between sets of points are perpendicular?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences