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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)
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)
=>\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=> 1-\(\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}=\frac{6}{6}+\frac{-1}{6}=\frac{5}{6}\)
Ta có :
Gọi A=1.2+2.3+3.4+4.5+...+49.50
\Rightarrow A=1.2+2.3+3.4+4.5+...+49.50
\Rightarrow 3.A=3.(1.2+2.3+3.4+4.5+...+49.50)
\Rightarrow 3.A=1.2.3+2.3.3+3.3.4+3.4.5+...+3.49.50
\Rightarrow 3.A=1.2.(3-0)+2.3.(3-0)+(3-0).3.4+(3-0).4.5+...+(3-0).49.50
\Rightarrow 3.A=1.2.3-0+2.3.3-0+3.3.4-0+3.4.5-0+...+3.49.50-0
\Rightarrow 3.A=1.2.3-0+2.3.4-1.2.3+5.3.4-2.3.4+...+49.50.51-48.49.50
\Rightarrow 3.A=49.50.51
\Rightarrow A=49.50.51:3
\Rightarrow A=49.50.17.3:3
\Rightarrow A=49.50.17
\Rightarrow A=41650
Đáp số : A=41650
3M = 1.2.3 + 2.3.(4-1) +..+ 99.100.(101-98)
3M = 1.2.3 + 2.3.4 - 1.2.3 + .... + 99.100.101 - 98.99.100
3M = 99.100.101 = 999900
M = 333300
sửa đề \(C=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=3\left(1-\frac{1}{2016}\right)=3.\frac{2015}{2016}=\frac{2015}{672}\)