SOLUTION: given any three consecutive integers prove that the product of the first and third number is one less than the squareof the middle one

SOLUTION: given any three consecutive integers prove that the product of the first and third number is one less than the squareof the middle one

Algebra.Com

Question 471539: given any three consecutive integers prove that the product of the first and third number is one less than the squareof the middle one
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Let x, x+1, x+2 be the three integers. The product of the first and third number is x(x+2) = x^2 + 2x. The square of the middle one is (x+1)^2 = x^2 + 2x + 1. Hence the product of the first and third numbers is one less than the square of the middle one.
RELATED QUESTIONS
for any three consecutive numbers prove that the product of the first and third numbers... (answered by Edwin McCravy)

find three consecutive integers such thAT the product of the first and the third is one... (answered by mananth)

find three consecutive even integers such that the product of the first and second is... (answered by CubeyThePenguin)

find three consecutive positive integers such that the product of the first and third one (answered by CubeyThePenguin)

Find the three consecutive odd integers such that the product of the first and third... (answered by ankor@dixie-net.com)

Find the three consecutive odd integers such that the product of the first and third... (answered by Mathtut)

Find three consecutive integers such that the product of the first and second is 36 less... (answered by ankor@dixie-net.com)

Find three consecutive positive odd integers such that twice the square of the first one (answered by jorel555)

find three consecutive odd integers such that three times the first integer is one less... (answered by peter_toribio90)