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\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)
4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)
4\(A=\frac{1}{6}-\frac{1}{406}\)
4\(A=\frac{100}{609}\)
\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)
=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406
=1/6-1/406
\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)
a)
\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)
\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)
b)
\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)
\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)
\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)
\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)
\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)
1,
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.........+\(\dfrac{1}{2^{2017}}\)
2B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\)
2B-B=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.......+\(\dfrac{1}{2^{2017}}\))
B=1-\(\dfrac{1}{2^{2017}}\)
Vậy B=1-\(\dfrac{1}{2^{2017}}\)
\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)
\(B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{ 1}{402}-\dfrac{1}{406}\)
\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}.\)