The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

maths: algebraic proof

prove that the sum of the squares of two consecutive odd numbers is always 2 more than a multiple of 8
Reply 1
Avatar for JSG29
JSG29
11
Let k be an integer. Then (2k+1)^2+(2k+3)^2 = 4k^2 + 4k + 1 + 4k^2 + 12k + 9 = 8(k^2 + 2k +1) + 2
Since k is an integer, (k^2 + 2k +1) is an integer, so 8(k^2 + 2k +1) is a multiple of 8 and thus the sum of the squares of 2 consecutive odd integers is 2 greater than a multiple of 8

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you