Lê Anh Duy
11/03/2018 at 12:58-
FC Alan Walker 11/03/2018 at 13:54
We have: \(A=2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)
\(=\left(2018-2017\right)\left(2018+2017\right)+\left(2016-2015\right)\left(2016+2015\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=2018+2017+2016+2015+...+2+1\)
\(=\dfrac{2018.2019}{2}=2037171\).
So the last two digits of A is 71.
Lê Anh Duy selected this answer.