Giải giúp mình với
E=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+\(\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+.....+24}\)
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\(\Rightarrow A=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+\frac{2}{\left(4+1\right).4}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
Vậy A=1.
Cái này có trong violympic vòng 10..bạn nhớ ôn cho kĩ nếu như bạn thi violympic!
Bạn có thể cho mình biết cách giải được không vậy bạn.
\(ĐặtA=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2A=1-\frac{1}{3^{2005}}\)
\(A=\frac{1-\frac{1}{3^{2005}}}{2}\)
Ủng hộ mk nha ^_-
Ta có:
\(E=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+....+\frac{1}{\frac{\left(24+1\right).24}{2}}\)
\(\Rightarrow E=\frac{2}{\left(2+1\right)2}+\frac{2}{\left(3+1\right)3}+...+\frac{2}{\left(24+1\right).24}\)
\(\Rightarrow E=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{24.25}\)
\(\Rightarrow E=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{25}\right)\)
\(\Rightarrow E=2.\left(\frac{1}{2}-\frac{1}{25}\right)=\frac{2.23}{50}=\frac{23}{25}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+24}\)
\(E=\frac{1}{3\cdot2:2}+\frac{1}{4\cdot2:2}+\frac{1}{5\cdot4:2}+...+\frac{1}{25\cdot24:2}\)
\(E=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{24\cdot25}\)
\(E=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{24}-\frac{2}{25}\)
\(E=1-\frac{2}{25}=\frac{23}{25}\)