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10 tháng 9 2018
M = 1 . 2 + 2 . 3 + ... + 2002 . 2003
3M = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + ... + 2002 . 2003 . ( 2004 - 2001 )
3M = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + ... + 2002 . 2003 . 2004 - 2001 . 2002 . 2003
3M = 2002 . 2003 . 2004
3M = 8036052024
M = 2678684008
19 tháng 7 2016
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 = \(\frac{2002.2003.2004}{3}=2678684008\)
JL
2
HP
13 tháng 3 2016
\(\frac{2000}{1.2}+...+\frac{2000}{2002.2003}\)
\(=2000.\left(\frac{1}{1.2}+....+\frac{1}{2002.2003}\right)\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2002}-\frac{1}{2003}\right)
\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2003}\right)=2000.\frac{2002}{2003}\)
13 tháng 3 2016
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
10 tháng 6 2020
*)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}\)
\(=\frac{5}{6}\)
*)\(\frac{2003}{1.2}+\frac{2003}{2.3}+\frac{2003}{3.4}+...+\frac{2003}{2002.2003}\)
\(=\frac{2003}{1}-\frac{2003}{2}+\frac{2003}{2}-\frac{2003}{3}+\frac{2003}{3}-\frac{2003}{4}+...+\frac{2003}{2002}-\frac{2003}{2003}\)
\(=2003-1\)
\(=2002\)
DN
0
TV
2
26 tháng 9 2020
\(M=1.2+2.3+3.4+...+2002.2003\)
\(3.M=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2002.2003.\left(2004-2001\right)\)
\(3.M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-...+2002.2003.2004-2001.2002.2003\)
\(3.M=2002.2003.2004\)
\(M=2002.2003.2004:3=2002.2003.668\)
\(M=2678684008\)
TA
26 tháng 9 2020
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
31 tháng 1
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 1
(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)
(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1 - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))
\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = \(\dfrac{1}{2003}\)
\(x\) = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)
\(x\) = \(\dfrac{2022}{2003}\)
\(\frac{2003}{1\cdot2}+\frac{2003}{2\cdot3}+...+\frac{2003}{2002\cdot2003}\)
\(=2003\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\right)\)
\(=2003\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=2003\cdot\left(1-\frac{1}{2003}\right)\)
\(=2003\cdot\frac{2002}{2003}\)
\(=\frac{2003\cdot2002}{2003}\)
\(=2002\)