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\(E=2\times4+4\times6+6\times8+...+98\times100\)
\(6\times E=2\times4\times6+4\times6\times\left(8-2\right)+6\times8\times\left(10-4\right)+...+98\times100\times\left(102-96\right)\)
\(=2\times4\times6+4\times6\times8-2\times4\times6+...+98\times100\times102-96\times98\times100\)
\(=98\times100\times102\)
\(\Rightarrow E=\frac{98\times100\times102}{6}=166600\)
Đặt:A = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))
=> A = 2.\(\frac{502}{1005}\)
=> A = \(\frac{1004}{1005}\)
đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\)
=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}\)
=\(\frac{502}{1005}\)
Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)
S=(2+98)*(4+6)+...+100+100+102
100*10+....+100+100*102
=224400
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)
\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)