1/2+1/(2^2)+1/(2^3)+...+1/(2^2019)
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TA
31 tháng 3 2023
Đặt D = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ...... + \(\dfrac{1}{2^{2019}}\)
⇔ 2D = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ...... + \(\dfrac{1}{2^{2018}}\)
⇔ D = 1 - \(\dfrac{1}{2^{2019}}\)
⇒ A = (1 - \(\dfrac{1}{2^{2019}}\)) : (1 - \(\dfrac{1}{2^{2019}}\))
⇒ A = 1
I
0
DT
17 tháng 7 2019
1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)
\(\Rightarrow1+2019^2=2020^2-2.2019\)
\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)
\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)
\(=2020\)
Vậy M=2020.
2) Xét : \(k\in N;k\ge2\)ta có:
\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)
\(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)
\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)
Cho \(k=3,4,...,2020.\)Ta có:
\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)
Vậy \(N=2018\frac{1009}{2020}.\)
HI
5
3 tháng 5 2018
=(2015/ 2019 + 3/2019 + 1/2019 ) : 1/2
= 2019/2019 x 2
= 1 x2
=2
3 tháng 5 2018
2015/2019:1/2+3/2019:1/2+1/2019:1/2
=(2015/2019+3/2019+1/2019):1/2
=1:1/2
=2
k cho mink nha
BL
0
BL
1
B
8 tháng 2 2020
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
---------------------
Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
Đặt A = 1/2 + 1/( 2^2 ) + 1/ ( 2 ^3 ) + ......+ 1 / ( 2 ^ 2019 )
=> 2A = 1 + 1/ 2 + 1/ ( 2 ^2 ) + ...... + 1/ ( 2 ^ 2018 )
- A = 1/2 + 1/ ( 2 ^ 2 ) + 1/ ( 2 ^ 3 ) + ...... + 1 / ( 2 ^ 2019 )
=> A = 1 - 1 / ( 2 ^ 2019 )
Vậy A = 1 - 1 / ( 2 ^ 2019 )
# có gì không hiểu thì cứ ib cho tui