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26 tháng 7 2015

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!

2 tháng 3 2020

sai nha

25 tháng 7 2015

\(A

16 tháng 5 2017

\(A=\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{4n-2}}-\frac{1}{3^{4n}}+....+\frac{1}{3^{98}}+\frac{1}{3^{100}}\)

Suy ra \(3^2.A=1-\frac{1}{3^2}+.....+\frac{1}{3^{4n-4}}-\frac{1}{3^{4n-2}}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\)

Khi đó \(3^2.A-A=1-\frac{1}{3^{100}}\)hay \(8A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1}{8}-\frac{1}{3^{100}}< 0,1\)

Vậy 

2 tháng 4 2018

2+12345678-5=

Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

14 tháng 7 2017

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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