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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:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)
\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)
\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)
\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)
\(A=\frac{100\cdot2}{1\cdot101}\)
\(A=\frac{200}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+...+\frac{4}{99\cdot101}-x-\frac{200}{101}=1\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)-x=1+\frac{200}{101}\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{101}\right)-x=\frac{301}{101}\)
\(\frac{4}{2}\cdot\frac{100}{101}-x=\frac{301}{101}\)
\(\frac{200}{101}-x=\frac{301}{101}\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)
Ta có : \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}-x-\frac{200}{101}=1\)
\(\Rightarrow\)\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=1+\frac{200}{101}+x\)
=> \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=\frac{301}{101}+x\)
=> \(2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2.\frac{100}{101}=\frac{301}{101}+x\)
=> \(\frac{200}{101}=\frac{301}{101}+x\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)