kiến thức lớp 8 chắc mới làm dc

\(A=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)+....+\frac{1}{\left(100-1\right)\left(100+1\right)}\)

\(A=\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)......\left(1+\frac{1}{100^2}\right)\)

ok tự giải típ nhé

A=(1+1/1.3)+........+(1+1/99.100)

=>A=[ (1.3+1)/(1.3 ) ] .[ (2.4+1)/(2.4) ] .... [ (99.101+1)/(99.101) ] 

=>A=( 4/1.3 ).( 9/2.4)......( 10000/99.101)

=>A=( 2²/1.3).( 3²/2..4).......( 100²/99.101)

=>A=\(\frac{2^2.3^2........99^2.100^2}{1.3.2.4.....99.101}\)

=>A=\(\frac{2.3....100.2.3.....100}{1.2.....99.3.4.....101}\)

=>A=\(\frac{100.2}{101}\)

=>A=\(\frac{200}{101}\)

Vậy A=\(\frac{200}{101}\)

\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{999x1001}\)

\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{1001-999}{999x1001}\)

\(2A=\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+...+\frac{1001}{999x1001}-\frac{999}{999x1001}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\)

\(2A=1-\frac{1}{1001}=\frac{1000}{1001}\)=> A = 500/1001

\(\frac{500}{1001}\)!

a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)

    Tổng A là: (2020+5)x404:2=409050

b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)

        \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

          \(=1-\frac{1}{101}=\frac{100}{101}\)

c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

         \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)

           \(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

             \(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)

Vậy .....

A = 5 + 10 + 15 + ... + 2015 + 2020

Số số hạng là : 404

A = 409050

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)

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

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

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

\(S=\frac{1}{2}\times\frac{58}{45}=\frac{29}{45}\)

Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$

Khi đó:

$1+\frac{1}{1.3}=\frac{2^2}{1.3}$

$1+\frac{1}{2.4}=\frac{3^2}{2.4}$

.........

$1+\frac{1}{99.101}=\frac{100^2}{99.101}$

Khi đó:

$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$

$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$

$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$

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