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LT
3
13 tháng 5 2016
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)
\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)
\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
....
\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
13 tháng 5 2016
\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)
=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)
=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)
=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
=\(\frac{5}{2}\cdot\frac{100}{101}\)
\(=\frac{250}{101}\)
TL
1
NQ
5
A
27 tháng 12 2015
2S=2/1.3+2/3.5+....+2/99.101
2S=1-1/3+1/3-1/5+....+1/99-1/101
2S=1-1/101
2S+1/101=1-1/101+1/101=1
Nho tick nha
27 tháng 12 2015
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(S=1-\frac{1}{101}=\frac{100}{101}\)
\(2S+\frac{1}{101}=\frac{100}{101}\)
\(S=2.\frac{100}{101}+\frac{1}{101}\)
\(\Rightarrow S=\frac{201}{101}\)
****
9 tháng 1 2022
\(S=1+\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{101}}\)
\(\Rightarrow S-1=\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{101}}\)
\(\Rightarrow\dfrac{1}{4}\left(S-1\right)=\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{103}}\)
\(\Rightarrow\dfrac{1}{4}\left(S-1\right)-\left(S-1\right)=\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{103}}-\dfrac{1}{2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{101}}\)
\(\Rightarrow\dfrac{3}{4}\left(S-1\right)=\dfrac{1}{2^{103}}\)
\(\Rightarrow S-1=\dfrac{1}{2^{103}}:\dfrac{3}{4}\)
\(\Rightarrow S-1=\dfrac{4}{3.2^{103}}\)
\(\Rightarrow S=\dfrac{4}{3.2^{103}}+1\)
DD
9 tháng 1 2022
S=1+12+123+125+...+12101S=1+12+123+125+...+12101
⇒S−1=12+123+125+...+12101⇒S−1=12+123+125+...+12101
⇒14(S−1)=123+125+127+...+12103⇒14(S−1)=123+125+127+...+12103
⇒14(S−1)−(S−1)=123+125+127+...+12103−12−123−...−12101⇒14(S−1)−(S−1)=123+125+127+...+12103−12−123−...−12101
⇒34(S−1)=12103⇒34(S−1)=12103
⇒S−1=12103:34⇒S−1=12103:34
⇒S−1=43.2103⇒S−1=43.2103
⇒S=43.2103+1
2 tháng 3 2017
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
2S = \(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}\)
\(2S=1-\frac{1}{101}\)
2S + 1/101 = \(1-\frac{1}{101}+\frac{1}{101}=1\)