NP
4
K
Khách
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QN
1
28 tháng 12 2018
\(A=2^{2018}-2^{2017}-2^{2016}-.....-2^1-2^0\)
\(\Rightarrow-A=2^{2018}+2^{2017}+2^{2017}+.....+2^1+2^0\)
\(\Rightarrow-A=2^0+2^1+2^2+......+2^{2017}+2^{2018}\)
\(\Rightarrow2\left(-A\right)=2+2^2+2^3+......+2^{2018}+2^{2019}\)
\(\Rightarrow2\left(-A\right)-\left(-A\right)=-A=2^{2019}-2^0\)
\(\Rightarrow A=-\left(2^{2019}-1\right)=-2^{2019}+1=1-2^{2019}\)
A
25 tháng 8 2021
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
25 tháng 8 2021
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
H
1
7 tháng 11 2017
Ax2=2+2^2+2^3+...+2^2018
Ax2 - A =(2+2^2+2^3+...+2^2018)-(2^0+2^1+2^2+...+2^2017)=2^2018-1
Mà 2^2018-1<2^2018 nên A<b
TN
1
CC
21 tháng 1 2020
Đặt biểu thức đã cho là A
Đặt \(B=2^{2019}+2^{2018}+.......+2^1+2^0\)
\(\Rightarrow2B=2^{2020}+2^{2019}+.......+2^2+2\)
\(\Rightarrow2B-B=B=2^{2020}-2^0\)
\(\Rightarrow A=2^{2020}-\left(2^{2020}-2^0\right)=2^{2020}-2^{2020}+1=1\)
18 tháng 7 2017
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{2018}\left(1+2+3+...+2018\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.\left(2+1\right)}{2}+\frac{1}{3}\cdot\frac{3.\left(3+1\right)}{2}+...+\frac{1}{2018}\cdot\frac{2018\left(2018+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2019}{2}\)
\(=\frac{2+3+4+...+2019}{2}\)
\(=\frac{\frac{2019\left(2019+1\right)}{2}-1}{2}=1019594.5\)
Ta có: \(A=2^0+2^1+2^2+2^3+...+2^{2018}\)
\(\Rightarrow2A=2^1+2^2+2^3+2^4+....+2^{2019}\)
\(\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{2019}\right)-\left(2^0+2^1+2^2+...+2^{2018}\right)\)
\(\Rightarrow A=2^{2019}-2^0=2^{2019}-1\)
Vậy ...
\(A=2^0+2^1+...+2^{2018}\)
\(\Rightarrow2A=2^1+2^2+...+2^{2019}\)
\(\Rightarrow2A-A=\left(2^1+2^2+...+2^{2019}\right)-\left(2^0+2^1+...+2^{2018}\right)\)
\(\Rightarrow A=2^{2019}-1\)