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ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
=>2B-B=\(1-\frac{1}{2^{99}}\)
mà 1/2^99>0 nên B<1 (đpcm)
C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}
\(\frac{C}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\frac{2C}{3}=C-\frac{C}{3}=\frac{1}{3}-\frac{1}{3^{100}}\)
\(2C=1-\frac{1}{3^{99}}\Rightarrow C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
C=.................................
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}\)
Dễ thấy \(1-\frac{1}{3^{99}}< 1\Leftrightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\Leftrightarrow C< \frac{1}{2}\)
