Question

[Nguyễn Hoàng An]

\(I=1+2-3-4+5+6-7-8+....+2009+2010\)

\(K=2011+2010\left(-4\cdot5^2+11+3^2\right)^{2009}\)

\(L=1+3-7+11-15+19\cdot23+...+99-103\)

                              help me !!!!!!!!!!!!!!

       Nhanh nhất 1tick

Answer 1

[Nguyễn Linh Chi]

\(I=\left(1+5+...+2009\right)+\left(2+6+...+2010\right)-\left(3+7+...+2007\right)-\left(4+8+...+2008\right)\)

\(=\frac{\left(2009+1\right).\left[\left(2009-1\right):4+1\right]}{2}+\frac{\left(2010+2\right)\left[\left(2010-2\right):4+1\right]}{2}\)

\(-\frac{\left(2007+3\right).\left[\left(2007-3\right):4+1\right]}{2}+\frac{\left(2008+4\right)\left[\left(2008-4\right):4+1\right]}{2}\)

\(=\frac{2010.503}{2}+\frac{2012.503}{2}-\frac{2010.502}{2}-\frac{2012.502}{2}\)

\(=1005+1006=2011\)