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Ta có : 1.2 + 2.3 + 3.4 + ... + 2008.2009
= ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 2008.2009.3 ) :3
= [ 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ... + 2008.2009.( 2010 - 2007 )] : 3
= [ 1.2.3 + 2.3.4 - 2.3.1 + 2.4.5 - 3.4.2 + ... + 2008.2009.2010 - 2008.2009.2007 ] : 3
= ( 2008.2009.2010 ) :3
= 2702828240
Đặt A = 1.2 + 2.3 + 3.4 + ..... + 2008.2009
<=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2008.2009.3
<=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ...... + 2008.2009.( 2010 - 2007 )
<=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2008.2009.2010 - 2007.2008.2009
<=> 3A = 2008.2009.2010
=> A = ( 2008.2009.2010 ) : 3
`S = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100.`
`3S = 1.2.3 + 2.3.(4-1) + 3.4.(5-4) + 4.5.(6-3) + ... + 99.100.(101-98)`
`3S = 1.2.3 + 2.3.4-1.2.3 + 3.4.5-4.5.6 + 4.5.6-3.4.5 + ... + 99.100.101-98.99.100`
`3S = 99.100.101`
`S = 33.100.101`
`S = 333300`
3S=1.2(3-0)+2.3(4-1)+.....+99.100(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-2.3.4+....+99.100.101-98-99-100
=99.100.101
S=33.100.101
=333300
=> 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
A = 1.2 + 2.3 + 3.4 + ....... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 99 . 100 . 3
3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +.... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 99 . 100 . 101 - 98 . 99 . 100
3A = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99 . 100 . 101
3A = 99 . 100 . 101 = 999900
A = 999900 : 3 = 333300
A=1*2+2*3+3*4+...+99*100
A=100*101*102:3
A=343400(công thức)
a=2009(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.........+1/2008-1/2009)
=2009x2008/2009
=2008
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101
=> 3S = 99.100.101
=> S = \(\frac{99.100.101}{3}=333300\)
ta xét
\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)
\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)
\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)
Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)
1.2 + 2.3 + 3.4 + 4.5 +...+ 2008.2009
= \(\frac{1}{3}\left(1.2.3+2.3.3+3.4.3+4.5.3+...+2008.2009.3\right)\)
= \(\frac{1}{3}\left(1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+2008.2009.\left(2010-2007\right)\right)\)
\(=\frac{1}{3}.2008.2009.2010=670.2008.2009\) số lớn nên bạn tự tính tiếp nhé!