A= \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{99}{100!}\).Hãy chứng minh A<1
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2 tháng 4 2020
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
=> \(3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
=> \(C+3C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> \(4C=1-\frac{100}{3^{100}}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
Đặt: \(B=-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
=> \(3B=-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
=> \(B+3B=-1-\frac{1}{3^{99}}\)
=> \(4B=-1-\frac{1}{3^{99}}\)
=> \(B=-\frac{1}{4}-\frac{1}{4}.\frac{1}{3^{99}}\)
=> \(4C=1-\frac{100}{3^{100}}+B=1-\frac{100}{3^{100}}-\frac{1}{4}-\frac{1}{4}.\frac{1}{3^{99}}\)
=> \(4C=\frac{3}{4}-\frac{100}{3^{100}}-\frac{1}{4.3^{99}}< \frac{3}{4}\)
=> \(C< \frac{3}{16}\)
13 tháng 7 2016
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
8 tháng 4 2018
a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
\(=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)\)
\(=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)
\(=\frac{16+4+1}{64}\)
\(=\frac{21}{64}< \frac{1}{3}\)(đpcm)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{100-1}{100!}=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+..+\frac{100}{100!}-\frac{1}{100!}\)
\(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{99!}-\frac{1}{100!}=1-\frac{1}{100!}<1\)
=> ĐPCM