Cho \(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2016}{^{3^{2016}}}\) CMR \(E< \frac{3}{4}\)
Những câu hỏi liên quan
Cho \(E=\frac{1}{3}+\frac{2}{3^2}-\frac{3}{3^3}+\frac{4}{3^4}-...+\frac{2016}{3^{2016}}-\frac{2017}{3^{2017}}\)
CMR : \(E< \frac{3}{16}\)
Cho \(E=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2015}{3^{2015}}-\frac{2016}{3^{2016}}\) . Chứng minh rằng \(E< \frac{3}{16}\)
Bài cuối đề thi học kỳ 2 môn toán trường mình đó , giải đi mk tk cho.
cho E=\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2015}{3^{2015}}-\frac{2016}{3^{2016}}\).Chứng minh rằng:E <\(\frac{3}{16}\)
Cho E = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{2015}{3^{2015}}-\frac{2016}{3^{2016}}\)
Chứng minh rằng :E < \(\frac{3}{16}\)
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
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Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017};B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\).CMR B/A là số nguyên
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
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Thuc hien phep tinh:
E=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
E = \(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\)- \(\frac{4}{3^4}\)+ .......+ \(\frac{2015}{3^{2015}}\) - \(\frac{2016}{3^{2016}}\) . Chứng minh E < \(\frac{3}{16}\)
CMR:
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2016^3}< \frac{1}{4}\)