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2 tháng 9 2020

\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)

\(2A-A=A\)

\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{49}}-\frac{1}{2^{50}}\)

\(=1-\frac{1}{2^{50}}< 1\)

\(\Rightarrow A< 1\)

             \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

          \(2A=\text{​​}\text{​​}1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

             \(A=1-\frac{1}{2^{50}}\)

             Vậy \(A\)<  1

22 tháng 9 2018

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

\(A=(\frac{-1.3}{2.2}).(\frac{-2.4}{3.3}).(\frac{-3.5}{4.4})...(\frac{-99.101}{100.100})\)

\(A=\frac{-1}{2}.\frac{101}{100}=\frac{-101}{200}<\frac{-100}{200}=\frac{-1}{2}\)

Vậy \(A<\frac{-1}{2}\)

_Học tốt_

9 tháng 5 2017
A=(-1.3/2.2)(-1.4/3.3)...(-99.101/100.100) A=-1/2.101/100=-101/200<-10/200=-1/2 vậy A<-1/2
16 tháng 1 2018

kaito kikru

10 tháng 9 2020

A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9999}{100^2}=-\frac{3.8.15...9999}{2.2.3.3.4.4...100.100}=-\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}\)

\(-\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=-\frac{1.101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=-\frac{1}{2}\)

=> A < - 1/2

10 tháng 9 2020

\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}-1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{100}+1\right)\)

Xét \(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\left(\frac{-1}{2}\right)\left(\frac{-2}{3}\right)\left(\frac{-3}{4}\right)...\left(\frac{-99}{100}\right)\)

Có 99 số hạng nhân với nhau nên kết quả cuối sẽ nhận dấu âm--->\(B=\frac{-1}{100}\)

Xét \(C=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{100}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{101}{100}=\frac{101}{2}\)

\(A=B.C=\frac{-1}{100}.\frac{101}{2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)