1x2+2x3+3x4+4x5+...2016x2017
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A = 1X2 +2x3 +...+ 2016x2107
3A = 1x2x3 + 2x3x3 + ...+ 2016x2017x3
3A = 1x2x(3-0) + 2x3x(4-1) + ... + 2016x2017x(2018-1)
3A = 1x2x3 - 1x2x0 +2x3x4 -1x2x3 +...+ 2016x2017x2018 - 2016x2017x2015
Ta loại trừ còn
3A = 2016x2017x2018 - 1x2x0
3A = 2016x2017x2018
A = 2016 x2017 x2018 : 3
A = 1x2 +2x3 +3x4 +...+ 2016 x 2017
3A = 1x2x3 + 2x3x3 +...+2016 x 2017 x3
3A = 1x2x(3-0) + 2x3x(4-1) +...+ 2016x2017x(2018-2015)
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\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2015\times2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}+\frac{1}{2016\cdot2017}\)
\(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{2016-2015}{2015\cdot2016}+\frac{2017-2016}{2016\cdot2017}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)(làm gọn một chút)
\(1-\frac{1}{2017}=\frac{2016}{2017}\)
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Gọi B = 1x2 + 2 x 3 + 3 x 4 + ... + 2016 x2017
3B = 3 x ( 1x2 + 2x3 + 3x4 + ... + 2016x2017)
= 1x2x3 + 2x3x3 + 3x4x3 + ... + 2016x2017x3 )
= 1x2x3 + 2x3x( 4-1) + 3x4x( 5 -2 ) + ... + 2016x2017x( 2018 - 2015)
= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2016x2017x2018 - 2015x2016x2017
= 2016 x2017 x2018
B = 672 x2017 x2018
Mà A = \(\frac{672x2017x2018}{2017x2018}\)
= 672
Vậy A = 672
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Đặt A=1x2+2x3+3x4+...+2016x2017
=>3A=3x1x2+3x2x3+3x3x4+...+3x2016x2017
=>3A=(3-0)x1x2+(4-1)x2x3+(5-2)x3x4+...+(2018-2015)x2016x2017
=>3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+...+2016x2017x2018-2015x2016x2017
=>3A=2016x2017x2018
=>A=\(\frac{2016\times2017\times2018}{3}\)(tự tính nha)
S = 1x2 + 2x3 + 3x4 + 4x5 + ... + 2016x2017
3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2016x2017x(2018-2015)
3S = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2016x2017x2018 - 2015x2016x2017
3S = 2016x2017x2018
S = 1/3 x 2016x2017x2018.