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Đặt A = 1.2 + 2.3 + 3.4 + ... + 2019.2020
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2019.2020.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + 2019.2020.(2021 - 2018)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2019.2020.2021 - 2018.2019.2020
= 2019.2020.2021
=> A = 2019.2020.2021 : 3 = 2 747 468 660
A=1.2+2.3+3.4+.............+2019.2020
3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018)
3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021
3A=2019.2020.2021
A=\(\frac{2019.2020.2021}{3}\)
A=2747468660
Vậy A=2747468660
Chúc bn học tốt
\(A=1.2+2.3+3.4+.......+2019.2020\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+......+2019.2020.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.........+2019.2020.\left(2021-2018\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.......+2019.2020.2021-2018.2019.2020\)
\(=2019.2020.2021\)
\(\Rightarrow A=\frac{3A}{3}=\frac{2019.2020.2021}{3}=2747468660\)
Vậy \(A=2747468660\)
\(C=2^2+4^2+...+2010^2\)
\(=2\left(1+1\right)+4\left(3+1\right)+...+2010\left(2009+1\right)\)
\(=2+1.2+4+3.4+...+2010+2009.2010\)\
\(=\left(2+4+...+2010\right)+\left(1.2+3.4+...+2009.2010\right)\)
Đặt A = 2+4+...+2010 = \(\frac{\left(2010+2\right).1005}{2}=1011030\)
Đặt B=1.2+3.4+...+2009.2010
3B=1.2.3+3.4.3+...+2009.2010.3
3B=1.2(3-0)+3.4(5-2)+...+2009.2010(2011-2008)
3B=1.2.3-0.1.2+3.4.5-2.3.4+...+2009.2010.2011-2008.2009.2010
3B=2009.2010.2011
B=\(\frac{2009.2010.2011}{3}=2706866330\)
Thay A và B vào C ta có:
\(C=1011030+2706866330=2707877360\)
B=1.2+2.3+...+2010.2011
3B=1.2.3+2.3.3+...+2010.2011.3
3B=1.2.(3-0)+2.3.(4-1)+...+2010.2011.(2012-2009)
3B=1.2.3-0.1.2+2.3.4-1.2.3+...+2010.2011.2012-2009.2010.2011
3B=(1.2.3+2.3.4+...+2010.2011.2012)-(0.1.2+1.2.3+...+2009.2010.2011)
3B=2010.2011.2012-0.1.2
3B=2010.2011.2012
B=\(\frac{2010.2011.2012}{3}=2710908440\)
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)
b) \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
=> \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}\)
c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\frac{4}{21}=11\)
d.e) ktra lại đề
Ta có:
n = \(2^{2020}-2^{2019}-2^{2018}-...-2-1\)
=> 2n = \(2^{2021}-2^{2020}-2^{2019}-2^{2018}-...-2^2-2\)
=> 2n - n = \(2^{2021}-2^{2020}-2^{2020}+1\)
=> \(n=2^{2021}-2.2^{2020}+1=1\)
=> \(A=2018.1-2019.1+2020.1=2019\)
\(C=1-2+2^2-2^3+...-2^{2011}+2^{2012}\)
\(\Rightarrow2C=2-2^2+2^3-2^4+...-2^{2012}+2^{2013}\)
\(\Rightarrow3C=1+2^{2013}\)
\(\Rightarrow C=\frac{1+2^{2013}}{3}\)
Vậy
\(D=-2+2^2-2^3+2^4-...-2^{2019}+2^{2020}\)
\(\Rightarrow-2D=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(\Rightarrow-3D=-2^{2021}+2\)
\(\Leftrightarrow D=\frac{2^{2021}-2}{3}\)