Question

Tính tổng: A=2+4+6+8...+2020-1-3-5-7-...2009

Answer 1

\(A=2+4+6+...+2020-1-3-5-7-...-2009\)

\(=\left(2+4+6+...+2020\right)-\left(1+3+5+...+2009\right)\)

\(=\dfrac{\left(2020+2\right).\left(\dfrac{2020-2}{2}+1\right)}{2}-\dfrac{\left(2009+1\right)\left(\dfrac{2009-1}{2}+1\right)}{2}\)

\(=1021110-1010025=11085\)

Answer 2

\(A=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2010-2009\right)+2011+2012+...+2020\\ A=1+1+1+...+1+2011+2012+...+2021\\ A=1005+2011+2012+...+2020\\ A=1005+\left(2020+2011\right)\cdot10:2=1005+20155=21160\)