# Answer on Algebra Question for reon

Question #12452

prove that the difference of the squares of two consecutive natural numbers is equal to their sum

Expert's answer

Let the first number be n, and the next one n+1:

(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 = n + (n+1).

(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 = n + (n+1).

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