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\(\left(1\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{13}{60}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...\left(\frac{1}{98}-\frac{1}{99}\right)\)

Ta thấy        \(\frac{13}{60}>\frac{12}{60}=\frac{1}{5}\)

          \(\frac{1}{6}-\frac{1}{7}>0\)

          \(\frac{1}{8}-\frac{1}{9}>0\)

\(...\)\(\frac{1}{98}-\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(\left(2\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

\(=\frac{23}{60}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

Ta thấy   \(\frac{23}{60}< \frac{24}{60}=\frac{2}{5}\)

      \(\frac{1}{7}-\frac{1}{8}>0\)

     \(\frac{1}{9}-\frac{1}{10}>0\)

\(...\frac{1}{97}-\frac{1}{98}>0\)

                \(\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

a, Ta có:

\(\frac{1}{2^3}< \frac{1}{1\cdot2\cdot3};\frac{1}{3^3}< \frac{1}{2\cdot3\cdot4};\frac{1}{4^3}< \frac{1}{3\cdot4\cdot5};...;\frac{1}{n^3}< \frac{1}{\left[n-1\right]n\left[n+1\right]}\)

\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)

Đặt \(A'=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)

\(\Rightarrow\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{\left[n-1\right].n}-\frac{1}{n\left[n+1\right]}\)

\(\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{n\left[n+1\right]}=\frac{1}{2}-\frac{1}{n\left[n+1\right]}=\frac{1}{4}-\frac{1}{2n\left[n+1\right]}< \frac{1}{4}\)

Vậy \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}< \frac{1}{4}\Leftrightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)

b,

\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)

\(=\left[1+1+1+...+1\right]+\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

Đặt \(C'=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3C'=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)

\(\Rightarrow3C'-C'=\left[1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right]-\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=1-\frac{1}{3^{98}}\)

\(\Rightarrow C'=\frac{1-\frac{1}{3^{98}}}{2}< 1\)

\(\Rightarrow98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}< 98+1=99< 100\)

\(\Rightarrow\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)

c,

\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}\)

\(4D=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\)

\(4D-D=\left[5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\right]-\left[\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}+\frac{5}{4^{39}}\right]\)

\(3D=5-\frac{5}{4^{39}}\Leftrightarrow D=\frac{5-\frac{5}{4^{39}}}{3}< \frac{5}{3}\)

Vậy:...........

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Bài này giải ra dài lắm.

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1<B<2 đi cho lẹ

Ta có: 9A=1+1/3²+...+1/3⁹⁸

Vậy 10A=(1+1/3²+...+1/3⁹⁸) + (1/3²+1/3⁴+...+1/3¹⁰⁰)

10A=1+2(1/3²+1/3⁴+...+1/3⁹⁸)+1/3¹⁰⁰

Vậy 10A>1 suy ra A > 0,1 suy ra người ra đề đã đặt sai đề!

sai nha

sai từ cái đề bài lun

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