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LG
2
KN
6 tháng 4 2019
Ta có: \(A=\frac{2^{2017}+2}{2^{2017}+3}=1-\frac{1}{2^{2017}+3}\)
\(B=\frac{2^{2017}+1}{2^{2017}+2}=1-\frac{1}{2^{2017}+2}\)
Vì \(\frac{1}{2^{2017}+3}< \frac{1}{2^{2017}+2}\) nên \(1-\frac{1}{2^{2017}+3}>1-\frac{1}{2^{2017}+2}\)
hay A > B
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 \(S< 2\)
FG
2
PM
0
H
0
Đặt A = \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2017}{2^{2017}}\)
2A = \(1+1+\frac{3}{2^2}+...+\frac{2017}{2^{2016}}\)
2A - A = A = \(2-\frac{1}{2}+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2016}}\)
A = \(\frac{3}{2}+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2016}}\)
Đặt B = \(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
2B = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
2B - B = B = \(\frac{1}{2}-\frac{1}{2^{2016}}\)
\(\Rightarrow A=\frac{3}{2}+\left(\frac{1}{2}-\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2016}}< \frac{3}{2}+\frac{1}{2}=2\)
Vậy A < 2