A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\) CMR:A<\(\frac{1}{16}\)
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2 tháng 5 2020
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
<=> \(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
=> \(5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^5}+...+\frac{99}{5^{100}}\right)\)
=> \(4A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=> \(20A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
=> \(20A-4A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)
=> \(16A=1+\frac{99}{5^{100}}-\frac{100}{5^{99}}< 1\)
=> \(A< \frac{1}{16}< \frac{1}{6}\)
26 tháng 6 2020
\(A=\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)
\(\Leftrightarrow5A=\frac{1}{5}+\frac{2}{5^2}+......+\frac{99}{5^{99}}\)
\(\Leftrightarrow5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+....+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)
\(\Leftrightarrow4A=\frac{1}{5}+\frac{1}{5^2}+......+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt : \(H=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}\)
\(\Leftrightarrow5H=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\)
\(\Leftrightarrow5H-H=\left(1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)
\(\Leftrightarrow4H=1-\frac{1}{5^{99}}\)
\(\Leftrightarrow H=\frac{1}{4}-\frac{1}{4.5^{99}}< \frac{1}{4}\)
\(\Leftrightarrow4A< B< \frac{1}{4}\)
\(\Leftrightarrow A< \frac{1}{16}\left(đpcm\right)\)
Ta có : A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
=> 5A = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
=> 5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\right)\)
=> 4A \(=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=> 20A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
Lấy 20A trừ A ta có :
20A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)
16A = \(1-\frac{99}{5^{99}}+\frac{99}{5^{100}}=1+99\left(\frac{1}{5^{100}}-\frac{1}{5^{99}}\right)=1-\frac{99.4}{5^{100}}\)
=> A = \(\frac{1}{16}-\frac{99}{4.5^{100}}< \frac{1}{16}\left(\text{ĐPCM}\right)\)
Ta có :A=\(\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)
5A=\(\frac{1}{5}+\frac{2}{5^2}+.....+\frac{99}{5^{99}}\)
5A -A=\(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{99}{5^{99}}\right)\)-\(\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)
4A =\(\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt B=\(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^{99}}\)
5B=\(1+\frac{1}{5}+...+\frac{1}{5^{98}}\)
5B - B =\(\left(1+\frac{1}{5}+...+\frac{1}{5^{98}}\right)\)- \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)
4B =\(1-\frac{1}{5^{99}}\)
Ta có :4A = B -\(\frac{99}{5^{100}}\)
16A = 4B -\(\frac{4.99}{5^{100}}\)=\(1-\frac{1}{5^{99}}-\frac{4.99}{5^{100}}\)
A = \(\frac{1}{16}-\frac{1}{5^{99}.16}-\frac{99}{5^{100}.4}\)< \(\frac{1}{16}\)
Suy ra: A <\(\frac{1}{16}\)