Só sánh S với 2::
S = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2017}{2^{2017}}\)
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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 \(S< 2\)
NN
0
NN
So sánh S với 2 biết :
S = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2017}{2^{2017}}\)
0
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 11 2021
Xét khai triển:
\(\left(1+x\right)^{2017}=C_{2017}^0+xC_{2017}^1+x^2C_{2017}^2+...+x^{2017}C_{2017}^{2017}\)
Lấy tích phân 2 vế:
\(\int\limits^1_0\left(1+x\right)^{2017}=\int\limits^1_0\left(C_{2017}^0+xC_{2017}^1+...+x^{2017}C_{2017}^{2017}\right)\)
\(\Leftrightarrow\dfrac{2^{2018}-1}{2018}=C_{2017}^0+\dfrac{1}{2}C_{2017}^1+...+\dfrac{1}{2018}C_{2017}^{2017}\)
Vậy \(S=\dfrac{2^{2018}-1}{2018}\)
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
