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26 tháng 9 2017

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\)

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}\\ \Rightarrow2A=1-\dfrac{1}{3^{2014}}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{2014}}< \dfrac{1}{2}\)

26 tháng 9 2017

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\)

=>\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\right)\)

=>\(2A=1-\frac{1}{2^{2014}}< 1\Rightarrow A< \frac{1}{2}\)(đpcm)

19 tháng 10 2016

Chứng minh rổng quát, Nếu:

\(A=\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+...+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\) (a;b \(\in\) N*)

\(a^{2.k}.A=1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+...+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\)

\(a^{2.k}.A+A=\left(1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+..+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\right)-\left(\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+..+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\right)\)

\(A.\left(a^{2.k}+1\right)=1-\frac{1}{a^{2.\left(k+n+1\right)}}< 1\)

\(A< \frac{1}{a^{2.k}+1}\)

Áp dụng vào bài toán dễ thấy a = 3; k = 1

Như vậy, \(A< \frac{1}{3^{2.1}+1}=\frac{1}{3^2+1}=\frac{1}{9+1}=\frac{1}{10}=0,1\left(đpcm\right)\)

20 tháng 10 2016

\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{2014}}-\frac{1}{3^{2016}}\)

\(\Rightarrow9A=1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{2012}}-\frac{1}{3^{2014}}\)

\(\Rightarrow10A=1-\frac{1}{3^{2016}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{2016}}}{10}\)

Vì 0,1 = \(\frac{1}{10}\) nên \(\frac{1-\frac{1}{3^{2016}}}{10}< \frac{1}{10}\) hay A < 0,1

NV
5 tháng 11 2019

\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}+\frac{1}{5^{2014}}\)

\(5M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)

\(\Rightarrow4M=1-\frac{1}{5^{2014}}< 1\)

\(\Rightarrow M< \frac{1}{4}< \frac{1}{3}\)

26 tháng 8 2016

Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2014^2}\)

Đặt A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{2014.2015}\)

\(A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.....\left(\frac{1}{2014}-\frac{1}{2015}\right)\\ =>A=\frac{1}{2}-\frac{1}{2015}\\ =>A=\frac{2013}{4030}\)

Mà S>A =>S>\(\frac{2013}{4030}\)

9 tháng 12 2019

Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

Tương tự : \(\frac{1}{3^2}< \frac{1}{2.3}\)\(\frac{1}{4^2}< \frac{1}{3.4}\); ......... ; \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2013.2014}\)               

        \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}\)

        \(=1-\frac{1}{2014}=\frac{2013}{2014}\)

\(\Rightarrow S< \frac{2013}{2014}\left(đpcm\right)\)

9 tháng 6 2017

sửa đề câu 1 :

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=1-\frac{1}{100!}< 1\)

sửa đề câu 2

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

20 tháng 6 2019

khi cộng cac số có tử bé hơn mẫu thì tổng sẽ <1 nha