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The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?


Which of the following equations is used in the process of solving this problem?

A3n^2 + 5 = 116
B3n^2 + 3n + 3 = 116
C3n^2 + 6n + 5 = 116

Verified Answer :

let's say, the first number is "n"

well, the consecutive number of "n" is "n + 1"and the consecutive number of "n+1" is "n + 1" +1 or n + 2

so, the numbers are (n), (n+1) and (n+2), whatever "n" is

now, the sum of their squares is 116

\(\bf (n)^2+(n+1)^2+(n+2)^2=116\)

expand the binomials by FOIL or binomial theorem, and then simplify

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