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22 tháng 4 2017

chứng tỏ :

Ta có : \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

áp dụng :

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

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

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

\(A=\frac{8}{9}\)

22 tháng 4 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(A=1-\frac{1}{9}=\frac{8}{9}\)

26 tháng 2 2018

a )  Ta có :   \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\) \(=\frac{1}{n.\left(n+1\right)}\)

b )   Áp dụng công thức trên tính tổng này như sau : 

             \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)

      \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

      \(=1-\frac{1}{10}\)

      \(=\frac{9}{10}\)

Chúc học giỏi !!!

26 tháng 2 2018

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

\(=\frac{1}{n\left(n+1\right)}=VT\RightarrowĐPCM\)

14 tháng 4 2016

Đặt \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=>3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\)

=> \(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{3n-1}-\frac{1}{3n+2}\)

=>\(3A=\frac{1}{2}-\frac{1}{3n+2}\)

=> \(3A=\frac{\left(3n+2\right):2}{3n+2}-\frac{1}{3n+2}\)

=> \(3A=\frac{1,5.n}{3n+2}\)

=>\(A=\frac{1,5.n}{3n+2}.\frac{1}{3}=>A=\frac{1,5.n}{\left(3n+2\right).3}=\frac{1,5.n}{9n+6}\)

\(Hay\) \(A=\frac{1,5n:1,5}{\left(9n+6\right):1,5}=\frac{n}{9n:1,5+6:1,5}=\frac{n}{6n + 4} \left(đpcm\right)\)

28 tháng 1 2016

Đặt A=1/2.5+1/5.8+...+1/(3n-1).(3n+2)

=>3A=3/2.5+3/5.8+...+3/(3n-1).(3n+2)

=>3A=1/2-1/5+1/5-1/8+...+1/3n-1-1/3n+2

=>3A=1/2-1/3n+2

=>3A=(3n+2-2)/[2.(3n+2)]

=>3A=3n/6n+4

=>A=3n/6n+4/3

=>A=n/6n+4

 

28 tháng 1 2016

210

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