Q = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

Q = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{2013.2015}\right)\)

Q =  \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

Q = \(\frac{1}{2}.\frac{2012}{6045}=\frac{1002}{6045}\)

\(Q=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\)

\(\Rightarrow Q.2=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\right)\)

\(\Rightarrow Q.2=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2013.2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{2012}{6045}\)

\(\Rightarrow Q=\frac{2012}{6045}.\frac{1}{2}=\frac{1006}{6045}\)

Mk tinh nhẩm, nên ko bt kết quả có đúng ko

nên bn thử tính lại kết quả nha!!!

Chúc bn hok tốt!!!

= 1/2. ( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +........+ 1/2013 - 1/2015)

= 1/2 . ( 1- 1/2015)

= 1007/2015

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{87.89}\)

= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{87}-\frac{1}{89}\right)\)

= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{89}\right)\)

= \(\frac{1}{2}.\frac{86}{267}=\frac{43}{267}\)

~~~
Đáp số to quá, tớ không chắc là mình đúng đâu.

#Sunrise

=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/87-1/89

=1/3-1/89

=86/267

trả lừoi

43/267

nha

hok tốt

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{13.15}\)

\(A=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

\(\frac{x}{2^2}\)+\(\frac{x}{2^3}\) +\(\frac{x}{2^4}\) =\(\frac{x}{3^2}\) +\(\frac{x}{3^3}\) +\(\frac{x}{3^4}\) là x =