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a=1.2+2.3+3.4+4.+....+200.201
3A = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 200.201.(202 - 199)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + .... + 200.201.202
3A = 200.201 . 202
A = 2706800
\(A=1.2+2.3+3.4+...+200.201\)
\(\frac{1}{A}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{201}=\frac{200}{201}\)
\(A=1:\frac{200}{201}=\frac{1.201}{200}=\frac{201}{200}\)
Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51
=> 3A = 49.50.51
= >A = 49.50.51/3 = 41650
3A=1.2.3+2.3.3+3.4.3+...+19.20.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
3A=19.20.21
=> \(A=\frac{19.20.21}{3}=2660\)
mk dùng cách của lớp 8 nha bạn ;
ta có công thức xích ma như sau x(x+1)
nhập vào xích ma ta có kết quả 2660
Ta có: 3A=1.2.3+2.3.3+3.4.3+.....+99.100.3
3A=1.2.3+2.3.(4-1)+3.4..(5-2)+....+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+99.100.101-98.99.100
3A=99.100.101
A=\(\frac{99.100.101}{3}\)
A=333300
Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A= 999900 : 3
A = 333300
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=9\left(1-\frac{1}{2020}\right)\)
\(=9.\frac{2019}{2020}\)
\(=\frac{18171}{2020}\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(A=9\left(1-\frac{1}{2020}\right)=\frac{9.2019}{2020}=\frac{18171}{2020}\)
...
Đặt S=1.2+2.3+.........+2011.2012
3S=1.2.3+2.3.(4-1)+...........+2011.2012.(2013-2010)
3S=1.2.3+2.3.4-1.2.3+...........+2011.2012.2013-2010.2011.2012
3S=2011.2012.2013
S=2011.2012.2013:3
S=2714954572
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2011.2012
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)
=> 3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012
=> 3A = 2011.2012.2013
=> A = \(\frac{2011.2012.2013}{3}=2714954572\).
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
\(S=1\cdot2+2\cdot3+3\cdot4+...+1000\cdot1001\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+1000\cdot1001\cdot\left(1002-999\right)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+1000\cdot1001\cdot1002-999\cdot1000\cdot1001\)
\(3S=1000\cdot1001\cdot1002\Rightarrow S=\frac{1000\cdot1001\cdot1002}{3}=334.334.000\)
A = 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = 33.100.101
A = 333300
ban gioi the