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ta có công thức 1.2+2.3+3.4+...+n.(n+1)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
áp dụng công thức vào bài ta có: 1.2+2.3+3.4+...+2002.2003 = \(\frac{2002.2003.2004}{3}=2678684008\)
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\(M=1.2+2.3+3.4+...+2002.2003\)
\(3.M=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2002.2003.\left(2004-2001\right)\)
\(3.M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-...+2002.2003.2004-2001.2002.2003\)
\(3.M=2002.2003.2004\)
\(M=2002.2003.2004:3=2002.2003.668\)
\(M=2678684008\)
M = 1 . 2 + 2 . 3 + 3 . 4 + ... + 2002 . 2003
3M = 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 3 + ... + 2002 . 2003 . 3
3M = 1 . 2 ( 4 - 1 ) + 2 . 4 ( 5 - 2 ) + 3 . 4 ( 6 - 3 ) + ... + 2002 . 2003 ( 2005 - 2002 )
3M = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + .... - 2002 . 2003 . 2004 + 2004 . 2005 . 2006
3M = 2005 . 2006 . 2007
3M = 2005 . 2006 . 889 . 3
M = 2005 . 2006 . 889
M = 4022030
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\(\frac{2000}{1.2}+...+\frac{2000}{2002.2003}\)
\(=2000.\left(\frac{1}{1.2}+....+\frac{1}{2002.2003}\right)\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2002}-\frac{1}{2003}\right)
\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2003}\right)=2000.\frac{2002}{2003}\)
đặt A=200/1.2+200/2.3+200/3.4+...+200/2002.2003
A:2000 = 1-1/2+1/2-1/3+...+1/2002-1/2003
A:2000=1-1/2003
A:2000=2002/2003
A=....
k nhe
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\(\frac{2003}{1\cdot2}+\frac{2003}{2\cdot3}+...+\frac{2003}{2002\cdot2003}\)
\(=2003\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\right)\)
\(=2003\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=2003\cdot\left(1-\frac{1}{2003}\right)\)
\(=2003\cdot\frac{2002}{2003}\)
\(=\frac{2003\cdot2002}{2003}\)
\(=2002\)
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3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3
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Ta có :
A = 1.2 + 2.3 + 3.4 + ... + 198.199 + 199.200
= 1.(1 + 1) + 2.(2 + 1) + 3.(3 + 1) + ... + 198(198 + 1) + 199(199 + 1)
= (1^2 + 1) + (2^2 + 2) + (3^2 + 3) + ... + (198^2 + 198) + (199^2 + 199)
= (1 + 2 + 3 + 4....+ 198 + 199) + (1^2 + 2^2 + 3^2 + ...+ 198^2 + 199^2)
* Dễ chứng minh :
....1 + 2 + 3 +...+ n = n(n + 1)/2
.... 1^2 + 2^2 +...+ n^2 = [n(n + 1)(2n + 1)]/6
Suy ra : A = [199.(199 + 1)]/2 + [199.(199 + 1)(2.199 + 1)]/6 = 2666600