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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)
- 225x17-225x16-225=225x[17-16-1]
=225x0
=0
- 28x76+24x28-2x48x18=28x[76+28] -2x48x18
= 28x104-1728
= 291-1728
=1184
\(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{2015\times2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\frac{1}{1}-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
Sửa đề nha :
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2016}{2017}=\frac{1008}{2017}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2016.2017}\)
\(=\frac{1}{2}\left[\left[\frac{1}{1}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{5}\right]+...+\left[\frac{1}{2016}-\frac{1}{2017}\right]\right]\)
= \(=\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\right]\)
\(=\frac{1}{2}.\left[1-\frac{1}{2017}\right]\)
= 1/2. 2016 / 2017 = 1008/2017
AI THẤY ĐÚNG ỦNG HỘ NHA
sửa đề \(C=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=3\left(1-\frac{1}{2016}\right)=3.\frac{2015}{2016}=\frac{2015}{672}\)
