1/4 + 1/28 1/70+....+1/103000
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Đặt :
\(M=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+................+\dfrac{1}{550}\)
\(\Rightarrow M=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+..................+\dfrac{1}{22.25}\)
\(\Rightarrow3M=\dfrac{3}{1.4}+\dfrac{3}{4.7}+.................+\dfrac{3}{22.25}\)
\(\Rightarrow3M=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.........+\dfrac{1}{22}-\dfrac{1}{25}\)
\(\Rightarrow3M=1-\dfrac{1}{25}\)
\(\Rightarrow3M=\dfrac{24}{25}\)
\(\Rightarrow M=\dfrac{24}{75}\)
Đặt:
\(A=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+.....+\dfrac{1}{550}\)
\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{22.25}\)
\(A=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{22}-\dfrac{1}{25}\right)\)
\(A=\dfrac{1}{3}\left(1-\dfrac{1}{25}\right)=\dfrac{1}{3}.\dfrac{24}{25}=\dfrac{25}{72}\)
Ta có:
1/4 = 1/3( 1-1/4)
1/28 = 1/3( 1/4 - 1/7)
1/70 = 1/3( 1/7 - 1/10)
..............................
1/10300 = 1/3( 1/100 - 1/103)
Cộng vế với vế ta có:
S = 1/4+1/28+1/70+1/130+...+1/10300 = 1/3( 1-1/103)
S = 34/103
\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)
\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(3M=1-\frac{1}{100}\)
\(3M=\frac{99}{100}\)
\(M=\frac{33}{100}\)
\(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+...+\dfrac{1}{10300}\\ =\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{100\times103}\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)
sửa đề: \(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+...+\dfrac{1}{10300}\)