1+\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...\frac{1}{300}\)
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\(S=\frac{1}{3}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{300}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{6}+\frac{1}{12}+\cdot\cdot\cdot+\frac{1}{600}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2\times3}+\frac{1}{3\times4}+\cdot\cdot\cdot+\frac{1}{24\times25}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{24}-\frac{1}{25}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{25}\)
\(\Rightarrow\frac{1}{2}S=\frac{23}{50}\)
\(\Rightarrow S=\frac{23}{50}:\frac{1}{2}\)
\(\Rightarrow S=\frac{23}{25}\)
S = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{300}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{600}\right)\)
= \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{24\times25}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{24}-\frac{1}{25}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{25}\right)\)
\(=2\times\frac{23}{50}\)
\(=\frac{23}{25}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{8}\right)\)
\(\Rightarrow A=2\cdot\frac{3}{8}=\frac{3}{4}\)
\(=1+\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+\frac{1}{5.3}+\frac{1}{3.7}+\frac{1}{7.4}+\frac{1}{4.9}+\frac{1}{9.5}\)
\(=1+1-\frac{1}{5}\)
\(=\frac{10}{5}-\frac{1}{5}\)
\(=\frac{9}{5}\)
Ai thấy đúng thì
lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)
= (4/50 + 3/70) + 2/100
= 7/120 + 2/100
= 9/220
p=1-1/2.5-1/3.5-1/1.3-1/4.7-1/2.3-1/3.7
p=1-(1/2.1/5-1/3.1/5)-(1/1.1/3-1/2.1/3)-(1/4.1/7-1/3.1/7)
p=1-(1/5.(1/2-1/3))-(1/3.(1-1/2))-(1/7.(1/4-1/3)
p=1-(1/5.1/6)-(1/3.1/2)-(1/7.-1/12)
p=1-1/30-1/6+1/84
p=341/420
\(S = \frac{1}{3} +\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28} \)
\(S=\frac{1}{3}+\frac{1}{3}.\frac{1}{2}+\frac{1}{5}.\frac{1}{2}+\frac{1}{5}.\frac{1}{3}+\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{4} \)
\(S=\frac{1}{3}(1+\frac{1}{2})+\frac{1}{5}(\frac{1}{2}+\frac{1}{3})+\frac{1}{7}(\frac{1}{3}+\frac{1}{4})\)
\(S=\frac{1}{3}.\frac{3}{2}+\frac{1}{5}.\frac{5}{6}+\frac{1}{7}.\frac{7}{12}\)
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\)
\(S=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}\)
\(S=\frac{9}{12}\)
\(S=\frac{3}{4}\)
Tính tổng hả bn
bn ơi đề bài là gì vậy