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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2015.2016}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
ta có \(3S=1\cdot2\cdot3+2\cdot3\cdot3+.....+99\cdot100\cdot3\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)....+99\cdot100\cdot\left(101-98\right)\)
\(3S=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-......-98\cdot99\cdot100+99\cdot100\cdot101\)
\(3S=99.100.101\)
\(S=\frac{99\cdot100\cdot101}{3}\)
S=...
3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
3S=99.100.101
S=33.100.101
S=333300
Vậy S=333300
A=1/2.3+1/3.4+1/4.5+..+1/2014.2015
A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2014-1/2015
A=1/2-1/2015
A=2013/4030
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
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
A=1/1-1/2+1/2-1/3+1/3-...........+1/2014-1/2015
A=1/1-1/2015
A=2014/2015
\(B=1.2+2.3+3.4+4.5+....+99.100\)
\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+99.100.101-98.99.100\)
\(3B=99.100.101\)
\(B=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)
Đặt S = 1.2 + 2.3 + 3.4 + ... + 2013.2014
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2013.2014.3
Mà :
1.2.3 = 1.2.3
2.3.3 = 2.3.4 - 2.3.1
3.4.3 = 3.4.5 - 3.4.2
...................................
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2012.2013.3 = 2012.2013.2014 - 2012.2013.2011
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012
Cộng tất cả, vế theo vế ---> 3S = 2013.2014.2015
---> S = 2013.2014.2015 / 3 = 2723058910
A=1.2+2.3+3.4+4.5+...+2014.2015
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2014.2015.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+2014.2015.(2016-2013)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2014.2015.2016-2013.2014.2015
=(1.2.3-1.2.3)+(2.3.4-2.3.4)+(3.4.5-3.4.5)+(4.5.6-4.5.6)+...+(2013.2014.2015-2013.2014.2015)+0.1.2+2014.2015.2016
=0+2014.2015.2016
=>A=\(\frac{2014.2015.2016}{3}\)