1. Tính C = \(\frac{1}{2}\)+ \(\frac{1}{2^3}\)+ \(\frac{1}{2^5}\)+ ... + \(\frac{1}{2^{99}}\)
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S
27 tháng 7 2019
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
22 tháng 11 2015
\(A=\frac{\frac{98}{2}+1+\frac{97}{3}+1+.....+\frac{2}{98}+1+\frac{1}{99}+1+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{99}+\frac{1}{100}}=\frac{\frac{100}{2}+\frac{100}{3}+........+\frac{100}{98}+\frac{100}{99}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{99}+\frac{1}{100}}\)
\(=\frac{100\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)}{\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)}=100\)
HN
25 tháng 3 2018
\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right).....\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-........-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-..........-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{99}{297}-\frac{3}{297}\right)\)
\(T=\frac{1}{2}.\frac{96}{297}\)
\(T=\frac{1}{2}.\frac{32}{99}\)
\(T=\frac{16}{99}\)
23 tháng 4 2020
\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right)\cdot.....\cdot\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(=\frac{1}{2\cdot3}\cdot\frac{3}{2\cdot5}\cdot\frac{5}{2\cdot7}\cdot.....\cdot\frac{97}{2\cdot99}\)
\(=\frac{1\cdot3\cdot5\cdot.....\cdot97}{2^{49}\left(3\cdot5\cdot7\cdot....\cdot99\right)}=\frac{1}{2^{49}\cdot99}\)
\(C=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2^2C=2+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}\)
\(\Rightarrow2^2C-C=2-\frac{1}{2^{99}}\)
\(3C=2-\frac{1}{2^{99}}\)
\(\Rightarrow C=\frac{2-\frac{1}{2^{99}}}{3}\)
\(C=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2^2C=2+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}\)
\(\Rightarrow4C-C=2-\frac{1}{2^{99}}\)
\(\Rightarrow C=\frac{2-\frac{1}{2^{99}}}{3}\)