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Khách
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29 tháng 11 2016
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
6 tháng 4 2021
Ta có: \(S=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+97\cdot98\cdot99\)
\(\Leftrightarrow4\cdot S=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+...+97\cdot98\cdot99\cdot\left(101-97\right)\)
\(\Leftrightarrow4\cdot S=98\cdot99\cdot100\cdot101\)
\(\Leftrightarrow S=\text{24497550}\)
QD
26 tháng 1 2017
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)
c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)
Câu b mình chưa nghĩ ra
Chúc bạn học tốt!
26 tháng 1 2017
a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)
= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)
b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)
= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))
= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?
c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).
NT
0
LH
1
PV
12 tháng 4 2016
Đặt S = 1/1.2.3 - 1/2.3.4 - 1/3.4.5 - ...- 1/97.98.99
S x 2 = 2/1.2.3 - 2/2.3.4 - 2/3.4.5 - ...- 2/97.98.99
= (1/1.2 -1/2.3) - (1/2.3 - 1/3.4 ) - (1/3.4 - 1/4.5) - ...- (1/97.98 - 1/98.99)
= 1/1.2 - 1/2.3 - 1/2.3 + 1/3.4 - 1/3.4 + 1/4.5 - ....- 1/97.98 + 1/98.99
= 1/2 -1/3 + 1/98.99
= 1618/9072 => S = 1618/9072 : 2 = 809/9072
11 tháng 11 2016
4(1.2.3) = 1.2.3.4 - 0.1.2.3
4(2.3.4) = 2.3.4.5 - 1.2.3.4
4(3.4.5) = 3.4.5.6 - 2.3.4.5
....................................
4(n-1)n(n+1) = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1)
=> 4 B = (n-1)n(n+1)(n+2) => B= (n-1)n(n+1)(n+2):4
11 tháng 11 2016
4(1.2.3)=1.2.3.4 - 0.1.2.3
4(2.3.4)=2.3.4.5 - 1.2.3.4
4(3.4.5)=3.4.5.6 - 2.3.4.5
.........................
......................................
.......................................
9 tháng 3 2015
Đặt A=1/1.2.3+1/2.3.4+...+1/99.100.101
2A=2/1.2.3+2/2.3.4+...2/99.100.101
2A=3-1/1.2.3+4-2/2.3.4+...+101-99/99.100.101
2A=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+101/99.100.101-99/99.100.101
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/99.100-1/100.101
2A=1/2-1/10100
VT
2
18 tháng 3 2016
=\(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
18 tháng 3 2016
4B = 1.2.3.4 + 2.3.4.4 + ... + (n-1)n(n+1).4
= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n-1)n(n+1)(n+2) - [(n-2)(n-1)n(n+1)]
= (n-1)n(n+1)(n+2) - 0.1.2.3
= (n-1)n(n+1)(n+2)
suy ra \(B = {(n-1)n(n+1)(n+2)\over 4}\)
24 tháng 12 2016
A=1.2.3+2.3.4+3.4.5+...+112.113.114
4A= 1.2.3.4+ 3.4.5.4+....+112.113.114.4
4A= 1.2.3.(4-0)+3.4.5.(6-2)+.....+ 112.113.114.(115-111)
4A= 1.2.3.4 - 0.1.2.3 + 3.4.5.6 - 2.3.4.5 +.....+ 112.113.114.115 - 111.112.113.114
4A= 112.113.114.115=165920160
A=165920160:4=41480040
chắc chắn 100% đó
24 tháng 12 2016
1.2.3+2.3.4+3.4.5+.......+112.113.114=1.2.3.4+2.3.4.4+........+112.113.114.4
=1.2.3.(5-1)+2.3.4.(6-1)+.................+112.113.114.(115-111)
=1.2.3-2.3.4+2.3.4-...........................-111.112.113+113.114.115
=113.114.115/3101
nho k minh nha
\(A=1\cdot2\cdot3+2\cdot3\cdot4+...+1998\cdot1999\cdot2000\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+....+1998\cdot1999\cdot2000\cdot4\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+....+1998\cdot1999\cdot2000\left(2001-1997\right)\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4-0\cdot1\cdot2\cdot3+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+1998\cdot1999\cdot2000\cdot2001-1997\cdot1998\cdot1999\cdot2000\)\(\Rightarrow4A=1997\cdot1998\cdot1999\cdot2000\)
\(\Rightarrow A=\frac{1997\cdot1998\cdot1999\cdot2000}{4}\)