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Giải
\(A=\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}\)
\(\Rightarrow A< \frac{1}{100}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
\(\Rightarrow A< \frac{100}{100}=1\)
Vậy A < 1 (đpcm)
A= 1/100x100+1/101x101+..........+1/199x199
Vì 1/100x100<99x100
1/101x101<100x101
...........
1/199x199 < 1/198x199
=) A< 1/99x100+1/100x101+...+1/198x199
A<1/99-1/100+1/100-1/101+.....+1/198-199
A<100/19701=0,0050....
Mà 1/100=0,01
=> A<1/100
K đúng nhé
Đặt \(S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(S=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có đpcm
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(>\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{51}\right)>\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{9}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{4}.\frac{3}{2}=\frac{3}{8}\)
\(\Rightarrow A>\frac{3}{8}\left(đpcm\right)\)