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A= 1.2 + 2.3 + 3.4 +...+ 2011.2012
=>3A= 1.2.3 + 2.3.3 + 3.4.3 +...+ 2011.2012.3
3A= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 2011.2012.(2013 - 2010)
3A= (1.2.3 + 2.3.4 + 3.4.5 +...+ 2011.2012.2013) - (0.1.2 + 1.2.3 + 2.3.4 +...+ 2010.2011.2012)
3A= (2011.2012.2013) - (0.1.2)
3A= 8144863716 - 0
\(A=\frac{8144863716}{3}=2714954572\)
Vậy A= 2714954572
\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)
\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)
\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)
3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=9\left(1-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2011.2012.3
=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 2011.2012.( 2013 - 2010 )
=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2011.2012.2013 - 2010.2011.2012
=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2010.2011.2012 - 2010.2011.2012 ) + 2011.2012.2013
=> 3S = 2011.2012.2013
=> S = ( 2011.2012.2013 ) : 3
3S=1.2.3+2.3.(4-1)+...............+2011.2012.(2013-2010)
3S=1.2.3+2.3.4-1.2.3+...............+2011.2012.2013-2010.2011.2012
3S=2011.2012.2013
S=2011.2012.2013:3
S=2714954572
\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2021.2022}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2021.2022}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\\ =4\left(1-\dfrac{1}{2022}\right)\\ =4.\dfrac{2021}{2022}\\ =\dfrac{4042}{1011}\)
4/1.2 4/2.3 4/3.4 ... 4/2021.4/2022
= 1/4. (1/1- 1/2+ 1/2- 1/3+ 1/3- 1/4+...+1/2021- 1/2022)
=1/4. (1/1- 1/2022)= 1/4. (2022/2022- 1/2022)
= 1/4. 2021/2022
= 2021/8088
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9\times\frac{99}{100}\)
\(A=\frac{891}{100}\) hoặc =8,91
A=9/1.2+9/2.3+9/3.4+...+9/98.99+9/99.100
A=9.(1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100)
A=9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)
A=9.(1/1-1/100)
A=9.99/100
A=891/100
A=8+91/100 ( viết dưới dạng hỗn số )
Vậy A=8+91/100
Nkớ k cho mink đó nha !!!
A = 1.2+2.3+3.4+.....+2011.2012
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 2011.2012.3
=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ........ + 2011.1012.(2013-2010)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........... + 2011.2012.2013 - 2010.2011.2012
=> 3A = (1.2.3 + 2.3.4 + 3.4.5 + ........ + 2011.2012.2013) - (1.2.3 + 2.3.4 + 3.4.5 + ........ + 2010.2011.2012)
=> 3A = 2011.2012.2013
=> A = \(\frac{2011\cdot2012\cdot2013}{3}\)