S=2(1-1/3+1/3-1/5+...+1/97-1/99)
  =2(1-1/99)
  =2(98/99)
  =196/99
 

2S=2/1*3+2/3*5+...+2/97*99

2S=1/1-1/3+1/3-1/5+...+1/97-1/99

2S=1-1/99

2S=98/99

S=49/99

\(S=7(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{63}) \)

\(S=7(\frac{1}{3}-\frac{1}{63})\)

\(S=7(\frac{21}{63}-\frac{1}{63}) \)

\(S=7.\frac{20}{63}\)

\(S=\frac{20}{9}\)

Do đó:\(S<\frac{5}{2}\)

S=\(\frac{2.7}{3.5}+\frac{2.7}{5.7}+\frac{2.7}{7.9}+....+\frac{2.7}{61.63}\)và\(\frac{5}{2}\)

S=7.(\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....-\frac{1}{63}\)) và\(\frac{5}{2}\)

S=7.(\(\frac{1}{3}-\frac{1}{63}\)) và\(\frac{5}{2}\)

S=7.\(\frac{20}{63}\)và\(\frac{5}{2}\)

=>S=\(\frac{20}{9}\)so với \(\frac{5}{2}\)

=>S=\(\frac{40}{18}\)và\(\frac{45}{18}\)

=>S<\(\frac{5}{2}\)

gọi A=1/1*2*3+1/2*3*4+...+1/49*50*51

      2A=2(1/1*2*3+1/2*3*4+...+1/49*50*51)

       2A=2/1*2*3+2/2*3*4+...+2/49*50*51

       2A=1/1*2-1/2*3+1/2*3-1/3*4+...+1/49*50-1/50*51

      2A=1/2-1/2550

      2A=637/1275

      A=637/1275:2

      A=637/2550

qua bài trên ta có công thức \(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\)= \(\frac{1}{n\cdot\left(n+1\right)}\)-\(\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}\)

lộn công thức là 2/n*(n+1)*(n+2)=1/n*(n+1)-1/(n+1)*(n+2) cho tui xin lỗi

mà tick nhé

mn giúp

Này bạn làm sao để ra dấu phân số vậy

Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)

\(=\frac{4}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\left(\frac{11}{55}-\frac{1}{55}\right)=2.\frac{10}{55}=2.\frac{2}{11}=\frac{4}{11}\)

Có gì không hiểu bạn hỏi lại mình nhé! Chúc bạn học tốt!

Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{53.55}\)

Đặt C = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{53.55}\)

     \(\frac{1}{2}C=\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+....+\left(\frac{1}{53}-\frac{1}{55}\right)\)

      \(\frac{1}{2}C=\frac{1}{5}-\frac{1}{55}\)

      \(\frac{1}{2}C=\frac{2}{11}\)

\(C=\frac{2}{11}:\frac{1}{2}\)

Vậy C = \(\frac{4}{11}\)
Có gì sai thì mong bạn thông cảm

Ghi bị lộn 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{101.103}\)

\(=>A=\frac{3}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\right)\)

\(=>A=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)

\(=>A=\frac{3}{2}.\left(1-\frac{1}{103}\right)=\frac{3}{2}.\frac{102}{103}=\frac{153}{103}>1\) (vì 153>103)

Vậy A>1

sorry,dòng thứ 2 sửa lại:\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{3}{101.103}\right)\) nhé!

\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2015^2}{2014.2016}=\frac{\left(2.3.4......2015\right)}{\left(1.2.3......2014\right)}.\frac{\left(2.3.4.....2015\right)}{\left(3.4.5......2016\right)}=\frac{2015}{1}.\frac{2}{2016}=\frac{2015}{1008}\)

\(A=-\frac{1}{20}+-\frac{1}{30}+...+-\frac{1}{90}\)

   \(=-\frac{1}{4.5}+-\frac{1}{5.6}+...+-\frac{1}{9.10}\)

   \(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{5}\right)+\left(-\frac{1}{5}\right)-\left(-\frac{1}{6}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)

   \(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{10}\right)=-\frac{3}{20}\)

Vậy \(A=-\frac{3}{20}\)