GửiTNs tao cuồng:c/m Bfrac{1}{2}+frac{1}{2^2}+frac{3}{2^3}+....+frac{100}{2^{100}}2Ta có:2B1+frac{1}{2}+frac{3}{2^2}+....+frac{100}{2^{99}}Rightarrow2B-BB1+left(frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+....+frac{1}{2^{99}}right)-frac{100}{2^{100}}(*)c/m frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+...+frac{1}{2^{99}}1Đặt Afrac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+....+frac{1}{2^{99}}Rightarrow2A1+frac{1}{2}+frac{1}{2^2}+...+frac{1}{2^{98}}Rightarrow2A-Aleft(1+frac{1}{2}+frac{1}{2^2}+...+frac{1}{2^{98}}right)-lef...
Đọc tiếp
Gửi
TNs tao cuồng:c/m \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{3}{2^3}+....+\frac{100}{2^{100}}<2\)Ta có:\(2B=1+\frac{1}{2}+\frac{3}{2^2}+....+\frac{100}{2^{99}}\)\(\Rightarrow2B-B=B=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)(*)c/m \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}<1\)Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)\)\(\Rightarrow A=1-\frac{1}{2^{99}}<1\)do đó \(B=1+A-\frac{100}{2^{100}}\Rightarrow B<2-\frac{100}{2^{100}}<2\left(đpcm\right)\)
