Tính bằng cách hợp lý ( nếu có thể ) :
\(\text{1.2 + 2.3 + 3.4 + ... + 2015.2016}\)
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a: \(M=1\cdot2+2\cdot3+\cdots+2020\cdot2021\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+2020\left(2020+1\right)\)
\(=\left(1^2+2^2+\cdots+2020^2\right)+\left(1+2+\cdots+2020\right)\)
\(=\frac{2020\left(2020+1\right)\left(2\cdot2020+1\right)}{6}+\frac{2020\cdot2021}{2}\)
\(=\frac{2020\cdot2021\cdot4041+3\cdot2020\cdot2021}{6}=\frac{2020\cdot2021\cdot4044}{6}\)
\(=2020\cdot2021\cdot674=2751551080\)
b: \(N=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+2019\cdot2020\cdot2021\)
\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+2020\left(2020-1\right)\left(2020+1\right)\)
\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+2020\left(2020^2-1\right)\)
\(=\left(2^3+3^3+\cdots+2020^3\right)-\left(2+3+\cdots+2020\right)\)
\(=\left(1^3+2^3+\cdots+2020^3\right)-\left(1+2+3+\cdots2020\right)\)
\(=\left(1+2+\cdots+2020\right)^2-\left(1+2+3+\cdots+2020\right)\)
\(=\left(2020\cdot\frac{2021}{2}\right)^2-2020\cdot\frac{2021}{2}=\left(1010\cdot2021\right)^2-1010\cdot2021\)
Đặt A = 1.2 + 2.3 + 3.4 + ... +2015.2016
3A = 1.2.3 + 2.3.(4-1) + ... + 2015.2016.(2017-2014)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 2015.2016.2017 - 2014.2015.2016
3A = 2014.2015.2016
A = 2727117120
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
A= 1.2+2.3+3.4+...+2015.2016
3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016
3A=2015.2016.2017
3A=8193538080
A=8193538080:3
A=2731179360
3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + 2015.2016.3
=> 3A = 1.2.3 + 2.3.( 4 -1 ) + 3.4.( 5 - 2 ) + .... + 2015.2016.( 2017 - 2014 )
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2015.2016.2017 - 2014.2015.2016
=> 3A = 2015.2016.2017
=> A = \(\frac{2015.2016.2017}{3}\)
A=1.2+2.3+3.4+...+2015.2016
=> 3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3
=> 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)
=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+ 2015.2016.2017-2014.2015.2016
=> 3A=2015.2016.2017
=> A=\(\frac{2015.2016.2017}{3}=2731179360\)
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Đặt \(A=1.2+2.3+3.4+...+2015.2016\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016\)
\(\Rightarrow3A=2015.2016.2017\)
\(\Rightarrow A=2015.2016.2017:3\)
\(\Rightarrow A=2015.672.2017\)
Vậy \(A=2015.672.2017\)
1 . 2 + 2 . 3 + 3 . 4 + ... + 2015 . 2016
3M = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 2015 . 2016 . 3
3M = 1 . 2 ( 3 - 0 ) + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 2015 . 2016 ( 2017 - 2014 )
3M = ( 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4. 5 + ... + 2015 . 2016 . 2017 ) - ( 0 . 1 . 2 + 1 . 2 . 3 + 2 . 3 . 4 + ... + 2014 . 2015 . 2016 )
3M = 2015 . 2016 . 2017
M = \(\frac{2015.2016.2017}{3}\)
M = 2731179360
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2015.2016
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2015.2016.3
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2015.2016.(2017 - 2014)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2015.2016.2017 - 2014.2015.2016
=> 3A = 2015.2016.2017
=> A = 2015.2017.672
=> A = 2 731 179 360