Tính tổng \(S=3^1-2^1+3^2-2^2+3^3-2^3+...+3^{2019}-2^{2019}\)
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CU
11 tháng 2 2019
A = (-1)(-1)^2(-1)^3...(-1)^2019
A = (-1)^1+2+3+...+2019
A = (-1)^2039190
A = 1
S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020
4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4
4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)
4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019
4S = 2018.2019.2020.2021
S = 2018.2019.2020.2021 : 4 = ...
11 tháng 7 2019
\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)
\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)
\(S=\frac{\left(2019-1\right)}{2}=1009\)
11 tháng 7 2019
\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)
\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)
\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)
\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)
XO
6 tháng 12 2020
Ta có \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2018}{2^{2018}}+\frac{2019}{2^{2019}}\)
=> 2S = \(1+1+\frac{3}{2^2}+...+\frac{2018}{2^{2017}}+\frac{2019}{2^{2018}}\)
Khi đó 2S - S = \(\left(1+1+\frac{3}{2^2}+..+\frac{2018}{2^{2017}}+\frac{2019}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2018}{2^{2018}}+\frac{2^{2019}}{2019}\right)\)
=> S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}-\frac{2019}{2^{2019}}\)
Đặt P = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\)
=> 2P = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)
Khi đó 2P - P = \(\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\right)\)
P = \(2-\frac{1}{2^{2018}}\)
Thay P vào S
=> S = \(2-\frac{1}{2^{2018}}-\frac{2019}{2^{2019}}=2-\frac{2}{2^{2019}}-\frac{2019}{2^{2019}}=2-\frac{2021}{2^{2019}}< 2\)
Vậy S < 2
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2019
\(A=\frac{2^{2019}}{2^{2020}-1}=\frac{1}{2}\left(\frac{2^{2020}-1+1}{2^{2020}-1}\right)=\frac{1}{2}\left(1+\frac{1}{2^{2020}-1}\right)\)
\(B=\frac{3^{2019}}{3^{2020}-1}=\frac{1}{3}\left(1+\frac{1}{3^{2020}-1}\right)< \frac{1}{2}\left(1+\frac{1}{3^{2020}-1}\right)< \frac{1}{2}\left(1+\frac{1}{2^{2020}-1}\right)\)
\(\Rightarrow B< A\)
DT
0
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
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2020
\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)
\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)
\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)
\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)
\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)
\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)
\(S=3-2+3^2-2^2+3^3-2^3+...+3^{2019}-2^{2019}\)
\(=\left(3+3^2+3^3+...+3^{2019}\right)-\left(2+2^2+2^3+...+2^{2019}\right)\)
\(=B-C\)
\(B=3+3^2+3^3+...+3^{2019}\)
\(3B=3^2+3^3+3^4+...+3^{2020}\)
\(3B-B=\left(3^2+3^3+3^4+...+3^{2020}\right)-\left(3+3^2+3^3+...+3^{2019}\right)\)
\(2B=3^{2020}-3\)
\(B=\frac{3^{2020}-3}{2}\)
\(C=2+2^2+2^3+...+2^{2019}\)
\(2C=2^2+2^3+2^4+...+2^{2020}\)
\(2C-C=\left(2^2+2^3+2^4+...+2^{2020}\right)-\left(2+2^2+2^3+...+2^{2019}\right)\)
\(C=2^{2020}-2\)
\(\Rightarrow S=B-C=\frac{3^{2020}-3}{2}-\left(2^{2020}-2\right)\)
\(=\frac{3^{2020}-3}{2}-\frac{2.\left(2^{2020}-2\right)}{2}\)
\(=\frac{3^{2020}-3-2^{2021}+4}{2}\)
\(=\frac{3^{2020}-2^{2021}+1}{2}\)
Vậy \(S=\frac{3^{2020}-2^{2021}+1}{2}\)