Tìm giá trị của 1x2+2x3+3x4+...+20x21
ai giúp giải zùm ik
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\(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+\(\dfrac{1}{99\times100}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
Đặt A = 1×2 + 2×3 + 3×4 + ... + 20×21
3A = 1×2×3 + 2×3×(4-1) + 3×4×(5-2) + ... + 20×21×(22-19)
= 1×2×3 - 1×2×3 + 2×3×4 - 2×3×4 + 3×4×5 - ... - 19×20×21 + 20×21×22
= 20×21×22
A = 20×21×22 : 3
= 20×22×7
= 3080
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Gọi biểu thức trên là S, ta có :
S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101
S = 99x100x101 : 3
S = 333300
ta có\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)
\(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=4\left(1-\frac{1}{2015}\right)\)
\(=4\cdot\frac{2014}{2015}\)
\(=\frac{8056}{2015}\)
VẬY A=\(\frac{8056}{2015}\)
Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}=\frac{891}{100}\)
P=1x2+2x3+3x4+...+2017x2018
3P = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2017x2018x3
3P = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... +2017x2018x(2019-2016)
3P = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2017x2018x2019 - 2016x2017x2018
3P = 2017x2018x2019
P = 2017x2018x2019 : 3
P = 2739315938
P = 1x2+2x3+3x4+...+2017x2018
3xP = 1x2x3+2x3x3+3x4x3+...+2017x2018x3
3xP = 1x2x3+2x3x(4-1)+3x4x(5-2)+...+2017x2018x(2019-2016)
3xP = 1x2x3+2x3x4-2x3x1+3x4x5-3x4x2+...+2017x2018x2019-2017x2018x2016
3xP = 2017x2018x2019
3xP = 8217947814
P = 8217947814 : 3
P = 2739315938
Gọi biểu thức trên là S, ta có:
\(S=\)\(1\cdot2+2\cdot3+3\cdot4+...+20\cdot21\)
\(3xS=3x\left(1\cdot2+2\cdot3+3\cdot4+...+20\cdot21\right)\)
\(3xS=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+20\cdot21\cdot3\)
\(3xS=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+20\cdot21\cdot\left(22-19\right)\)
\(3xS=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+20\cdot21\cdot22-19\cdot20\cdot21\)
\(3xS=20\cdot21\cdot22\)
\(S=20\cdot21\cdot\frac{22}{3}\)
\(S=3080\)