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16 tháng 4 2019

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow A< 1-\frac{1}{2017}=\frac{2016}{2017}\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)

26 tháng 8 2020

Bài làm:

Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\)

=> \(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\right)\)

<=> \(2B=1-\frac{1}{3^{2017}}\)

=> \(B=\frac{1}{2}-\frac{1}{3^{2017}.2}< \frac{1}{2}\)

=> \(B< \frac{1}{2}\)

12 tháng 12 2017

Ta có :

\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\)

\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\)

\(2S-S=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\right]\)

\(S=1-\left(\frac{1}{2}\right)^{2017}< 1\)

24 tháng 11 2017

Ta có :

\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)

vì \(\frac{a1}{a2}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\) 

\(\frac{a2}{a3}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)

...

\(\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
\(\Rightarrow\frac{a1}{a2}.\frac{a2}{a3}.\frac{a3}{a4}...\frac{a2016}{a2017}=\frac{\left(a1+a2+a3+...+a2016\right)^{2016}}{\left(a2+a3+a4+...+a2017\right)^{2016}}\)

\(\Rightarrow\frac{a1}{a2017}=\left(\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\right)^{2016}\)

24 tháng 11 2017

Ta có a1/a2=a2/a3=a3/a4=...=a2016/a2017

=> a1/a2=(a1+a2+a3+...+a2016)

/(a2+a3+a4+...+a2017)

=> a12016/a22016 =(a1+a2+a3+...+a2016)2016 /(a2+a3+a4+...+a2017)2016 (1)

Ta lại có a1/a2=a2/a3=a3/a4=...=a2016/a2017

=> a12016/a22016= a1/a2.a2/a3.a3/a4. ... .a2016/a2017=a1/a2017 (2)

Từ (1) và (2) => đpcm

5 tháng 5 2017

\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\\ 2A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\\ 2A-A=\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\right)\\ A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{2017}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ A=1-\dfrac{4034}{2^{2017}}-\dfrac{2016}{2^{2017}}\\ A=1-\left(\dfrac{4034}{2^{2017}}+\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{6050}{2^{2017}}< 1\)

Vậy \(A< 1\)

8 tháng 3 2018

Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016

          A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016

           A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016

           A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016

           A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016

            A=2016  -  2014.(1/2015+1/2016+....+1/4030)   -2016

             A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)

             A=-2014.(1/2015+1/2016+....+1/4030)

   mà B = 1/2015+1/2016+....+1/4030

      nên A : B = -2014

8 tháng 3 2018

các bn hãy ủng hộ mk nhé !!! Thanks everyone!!!

18 tháng 8 2020

cảm ơn bạn nhiều