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\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)
< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)
ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
Ta có:
\(\hept{\begin{cases}\frac{1}{5}=\frac{1}{5}\\\frac{1}{16}< \frac{1}{5}\\\frac{1}{113}< \frac{1}{5}\end{cases}}...\)\(\Rightarrow\frac{1}{5}+\frac{1}{16}+\frac{1}{25}+\frac{1}{41}+\frac{1}{60}+\frac{1}{85}+\frac{1}{113}< \frac{1}{5}.7=\frac{7}{5}< \frac{10}{5}=2\)(ĐPCM)
Đặt \(S=\frac{1}{3}+\frac{2}{3^2}+.......+\frac{101}{3^{101}}\)
\(\Rightarrow3S=1+\frac{2}{3}+.......+\frac{101}{3^{100}}\)
\(\Rightarrow3S-S=\left(1+\frac{2}{3}+..+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2S=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+....+\frac{1}{3^{100}}\)
\(\Rightarrow6S< 3+1+........+\frac{1}{3^{99}}\)
\(\Rightarrow6S-2S< \left(3+1+....+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+....+\frac{1}{3^{100}}\right)\)
\(\Rightarrow4S< 3-\frac{1}{3^{100}}< 3\Rightarrow S< \frac{3}{4}\)
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{101}{3^{101}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\right)\)
\(4A=3-\frac{101}{3^{100}}-\frac{1}{3^{100}}+\frac{101}{3^{101}}\)
\(4A=3-\frac{303}{3^{101}}-\frac{3}{3^{101}}+\frac{100}{3^{101}}\)
\(4A=3-\frac{206}{3^{101}}< 3\)
=>\(4A< 3\)
\(\Rightarrow A< \frac{3}{4}\)
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