Tính: B=(1-1/(1+2)).(1-1/(1+2+3))...(1-1/(1+2+3+...+100))
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KT
29 tháng 7 2018
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow\)\(A=2-\frac{1}{2^{100}}\)
\(B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\)\(3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\)
\(\Rightarrow\)\(3B-B=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow\)\(2B=3-\frac{1}{3^{100}}\)
\(\Rightarrow\)\(B=\frac{3-\frac{1}{3^{100}}}{2}\)
26 tháng 3 2018
Ta có : \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{100}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=2-\frac{1}{2^{99}}\)
\(\Leftrightarrow A=2-\frac{1}{2^{99}}\)
B tương tự
DT
0
20 tháng 8 2016
1b) Ta có: \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{101}{100}=\frac{3.4.5....101}{2.3.4....100}=\frac{101}{2}\)
B=\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\).....\(\left(1-\frac{1}{1+2+3+4+...+100}\right)\)
B= \(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).....\)\(\left(1-\frac{1}{100.101:2}\right)\)
B= \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.....\frac{5049}{5050}\)
B= \(\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}.....\frac{5049.2}{5050.2}\)
B= \(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{99.102}{100.101}\)
B=\(\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{4.5.6.7.....102}{3.4.5.6.....101}\)
B=\(\frac{1}{100}.\frac{102}{3}=\frac{17}{50}\)