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8 tháng 3 2016

a)\(\Leftrightarrow\frac{1}{n\left(n+1\right)}=\frac{n+1-1}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)(đpcm)

b)\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{8}\)

\(\Rightarrow A=\frac{3}{8}\)

27 tháng 2 2016

\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)};\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

\(Vậy\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow A=1-\frac{1}{8}=\frac{7}{8}\)

27 tháng 2 2016

1a,Là điều hiển nhiên khỏi cần giải

b,=1-1/10

2,1/2-1/8

24 tháng 2 2018

mong các bạn giúp mình nhé

mình xin cảm ơn

30 tháng 3 2019

Biết câu b thôi, với lại k cần áp dụng câu a)

b. \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\) 

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

\(=1-\frac{1}{10}\) 

\(=\frac{9}{10}\)

11 tháng 5 2020

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}=\frac{5}{6}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)

16 tháng 3 2017

a) Vì n.(n+1) = 1/n-1/n+1 suy ra n thuộc N      n khác 0

b) A=1/1*2+1/2*3+1/3*4+...+1/9.10

A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

A=1-1/10=9/10

Vậy A = 9/10

5 tháng 5 2016

b) A=1/2.3+1/3.4+....+1/99.100

=> A=1/2-1/3+1/3-1/4+....+1/99-1/100

=> A=1/2-1/100

=> A=50/100-1/100

=> A=49/100

5 tháng 5 2016

49/100 

k nhe

7 tháng 6 2016

a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)

b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)

9 tháng 6 2017

b) 

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

\(B=\frac{1}{5}-\frac{1}{12}\)

\(B=\frac{7}{60}\)

9 tháng 6 2017

a) Ta có: 

\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)  ;   \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{1\left(n+1\right)}\)

Vậy \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)

b)  \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{100-99}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

\(B=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-\frac{1}{11}-\frac{1}{12}\)

\(B=\frac{1}{5}-\frac{1}{12}\)

\(B=\frac{7}{60}\)