\(R=\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+...+\frac{2006.2006}{2005.2007}\)

\(R=\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{2006^2}{2005.2007}\)

\(R=\frac{1.3+1}{1.3}+\frac{2.4+1}{2.4}+\frac{3.5+1}{3.5}+...+\frac{2005.2007+1}{2005.2007}\)

\(R=1+\frac{1}{1.3}+1+\frac{1}{2.4}+1+\frac{1}{3.5}+...+1+\frac{1}{2005.2007}\)  

\(R=\left(1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2005.2007}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2004.2006}\right)\)

( có 2005 số 1)

\(R=2005+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)

\(+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2004}-\frac{1}{2006}\right)\)

\(R=2005+\frac{1}{2}.\left(1-\frac{1}{2007}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2006}\right)\)

\(R=2005+\frac{1}{2}\cdot\frac{2006}{2007}+\frac{1}{2}\cdot\frac{501}{1003}\)

\(R=2005+\frac{1003}{2007}+\frac{501}{2006}\)

...

đến đây bn tự tính típ nha!

=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+............+\frac{1}{18.19.20}\)

=\(\frac{2}{1.2.3.2}+\frac{2}{2.3.4.2}+............+\frac{2}{18.19.20.2}\)

=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}............+\frac{1}{18.19}-\frac{1}{19.20}\)

=\(\frac{1}{1.2}-\frac{1}{19.20}\)

=\(\frac{189}{380}\)

\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)

\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)

\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)

\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)

\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)

\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)

\(\Rightarrow S=125,4372197\)

\(\)

thx  you

ai đó giúp tui làm đc ko? Ai làm xong trước sẽ đc k nha!!!!

à nhầm, là "k" chứ ko phải là "k" đâu!

Tóm lại là ai xong trước sẽ đc chấm đúng nha!

a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =100/101

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5

    =(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005