1/1x3 + 1/3x5 + ...+ 1/99x101 = ?
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5 tháng 3 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}\)
\(=\frac{50}{101}\)
5 tháng 3 2019
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)
\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
2
23 tháng 2 2017
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(=2\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right):2\)
\(=\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right):2\)
\(=\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{101}\right):2\)
\(=\frac{100}{101}:2=\frac{50}{101}\).
NT
2
AH
Akai Haruma
Giáo viên
15 tháng 8 2023
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:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
NK
4
MT
12 tháng 7 2015
\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)
\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)
\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)
\(=\frac{2\times100}{101}=\frac{200}{101}\)
TN
0
NP
2
DP
2 tháng 7 2017
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+...+\frac{4}{99\cdot101}-x-\frac{200}{101}=1\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)-x=1+\frac{200}{101}\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{101}\right)-x=\frac{301}{101}\)
\(\frac{4}{2}\cdot\frac{100}{101}-x=\frac{301}{101}\)
\(\frac{200}{101}-x=\frac{301}{101}\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)
2 tháng 7 2017
Ta có : \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}-x-\frac{200}{101}=1\)
\(\Rightarrow\)\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=1+\frac{200}{101}+x\)
=> \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=\frac{301}{101}+x\)
=> \(2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2.\frac{100}{101}=\frac{301}{101}+x\)
=> \(\frac{200}{101}=\frac{301}{101}+x\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)
TN
1
TB
17 tháng 1 2016
ta có : 2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
2S=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
2S=\(\frac{1}{1}-\frac{1}{101}\)
2S+\(\frac{1}{101}\)= \(\frac{1}{1}-\frac{1}{101}+\frac{1}{101}\)
2S+\(\frac{1}{101}\)=1
ok
DK
0