Tính: S = 1.2+2.3+3.4+4.5+...+99.110
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`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
S=1.2+ 2.3+4,5.......+99.100
Nhân cả 2 vế với 3, ta được:
3S=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
----> S = (99.100.101):3
S= 333300
Vậy A=333300
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
=> 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
S=1.2+2.3+3.4+...+99.100
3S=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3S=99.100.101
S=(99.100.101):3=333300
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 \(S\left(100\right)=\frac{99.100.101}{3}=333300\)