Cho S=\(\dfrac{2}{2^1}\)+\(\dfrac{3}{2^2}\)+\(\dfrac{4}{2^3}\)+...+\(\dfrac{n+1}{2^n}\)+...+\(\dfrac{2017}{2^{2016}}\)
So sánh S với 3.
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14 tháng 6 2021
Giải:
\(S=\dfrac{1}{2}+\dfrac{2}{2^2}+...+\dfrac{n}{2^n}+...+\dfrac{2017}{2^{2017}}\)
Với \(n>2\) thì \(\dfrac{n}{2^n}=\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\)
Ta có:
\(\dfrac{n+1}{2^{n-1}}=\dfrac{n+1}{2^n:2}=\dfrac{2.\left(n+1\right)}{2^n}\)
\(\Rightarrow\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\)
\(=\dfrac{2.\left(n+1\right)}{2^n}-\dfrac{n+2}{2^n}\)
\(=\dfrac{2.\left(n+1\right)-n-2}{2^n}\)
\(=\dfrac{n}{2^n}\)
\(\Leftrightarrow S=\dfrac{1}{2}+\left(\dfrac{2+1}{2^{2-1}}-\dfrac{2+2}{2^2}\right)+...+\left(\dfrac{2016+1}{2^{2015}}-\dfrac{2018}{2^{2016}}\right)+\left(\dfrac{2017+1}{2^{2016}}-\dfrac{2019}{2^{2017}}\right)\)
\(S=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{2019}{2017}\)
\(S=2-\dfrac{2019}{2017}\)
\(\Leftrightarrow S=2-\dfrac{2019}{2017}< 2\)
Hay \(S< 2\)
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)
\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)
\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)
\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)
Tới đây thì đơn giản rồi nhé