Tính tổng A=1.2+2.3+3.4+...+2013.2014
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Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
Tham khảo:
A=1.2+2.3+3.4+...+2013.2014
3A = 1.2.3 + 2.3.3 + 3.4.3 +...+ 2013.2014.3
Mà: 1.2.3 = 1.2.3
2.3.3 = 2.3.4 - 2.3.1
3.4.3 = 3.4.5 - 3.4.2
2012.2013.3 = 2012.2013.2014 - 2012.2013.2011
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012
=> 3S = 2013.2014.2015
=> A = 2013.2014.2015 / 3 = 2723058910
3A= 1.2.3+2.3.3+3.4.3+...........+2010.2011.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.........+2010.2011.(2012-2009)
=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+2010.2011.2012-2009.2010.2011
=>3A=2010.2011.2012
=>3A=3.670.2011.2012
=>A=670.2011.2012
=>A= .......lấy máy tính mà tính
E = 1.2+2.3+3.4+......+99.100
Gấp E lên 3 lần ta có:
E . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
E . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
E . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 E . 3 = 99.100.101
E = 99.100.101 : 3
E = 33.100.101
E = 333 300
k mik nha
E = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3E = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 99.100.(101-98)
=> 3E = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3E = 99.100.101
=> E = 333300
Đặt A = 1.2+2.3+3.4+....+98+99
ð 3a = 1.2.3-1.2.3+2.3.4+...+98.99.100
ð 3a=98.99.100
ð A=98.99.100/3
ð A=323400
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 98.99
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ....... + 98.99.100
=> 3A = 98 .99.100
=> A = 98 .99.100/3
=> A = 323400
Ta có : A = 1/1.2 + 1/2.3 + .... + 1/98.99 + 1/99.100 .
=> A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/98 - 1/99 + 1/99 - 1/100 .
=> A = 1 - 1/100 .
=> A = 99/100 .
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)
\(=\frac{1999}{1000}\)
Tham khảo nhé~
TL:
a)\(2+4+6+...+2000=\frac{\left(2+2000\right).\left[\left(2000-2\right):2+1\right]}{2}\)
\(=1001000\)
Câu b tương tự nha bạn:)
c) Đặt 1.2+2.3+....+99.100 =A
\(3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+...99.100.101-98.99.100\)
\(3A=99.100.101\)
\(A=333300\)
Vậy .....
a) Đặt A= 2+4+6+...+1998+2000
Ta có: A=(2+2000).1000:2
=> A=2002.1000:2
=> A=2002000:2
=> A=1001000
b) Đặt B= 5+9+13+...+1997+2001
=> B=(2001+5).500:2
=> B=2006.500:2
=> B=1003000:2
=> B=501500
c)Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 => 3S = 3.33.100.101
=> S=33.100.101= 333300