giải hộ với:
C= 1x2x3+2x3x4+...........+8x9x10
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28 tháng 11 2015
4S = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 - 1) + .... + 8 x 9 x 10 x (11 - 7)
4S = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + .... + 8 x 9 x 10 x 11 - 7 x 8 x 9 x 10
4S = (1 x 2 x 3 x 4 - 1 x 2 x 3 x 4) + ..... + (7 x 8 x 9 x 10 - 7 x 8 x 9 x 10) + 8 x 9 x 10 x 11
4S = 8 x 9 x 10 x 11 = 7920
S = 7920 : 4 = 1980
DN
6
8 tháng 8 2016
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{22}{45}\)
2 tháng 4 2017
Gọi tổng trên là S , ta có :
S = 1/1.2.3 + 1/2.3.4 + ... + 1/8.9.10
S.2 = 2/1.2.3 + 1/2.3.4 + ... + 1/8.9.10
S.2 = 3 -1 /1.2.3 + 4 - 2/2.3.4 + ... + 10 - 8/8.9.10
S.2= 3/1.2.3 - 1/1.2.3 + 4/2.3.4 - 2/2.3.4 + ... + 10/8.9.10 - 8 /8.9.10
S.2 =1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/8.9 - 1/9.10
S.2 = 1/2 - 1/90
S = 1/4 - 1/360
S= 89/360
GN
3
18 tháng 3 2018
\(A=1\cdot2\cdot3+2\cdot3\cdot4+...+7\cdot8\cdot9+8\cdot9\cdot10\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+7\cdot8\cdot9\cdot4+8\cdot9\cdot10\cdot4\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+7\cdot8\cdot9\cdot\left(10-6\right)+8\cdot9\cdot10\cdot\left(11-7\right)\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+7\cdot8\cdot9\cdot10-6\cdot7\cdot8\cdot9+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)
\(4A=8\cdot9\cdot10\cdot11\)
\(A=\frac{8\cdot9\cdot10\cdot11}{4}=1980\)
HT
2
NN
28 tháng 1 2021
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{8.9.10}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+......+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
NT
0
MT
1
31 tháng 1 2021
Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)
\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)
hay \(A=\dfrac{-4949}{19800}\)
NN
10 tháng 6 2020
\(C=1.2.3+2.3.4+........+48.49.50\)
\(\Rightarrow4C=1.2.3.4+2.3.4.4+........+48.49.50.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+.........+48.49.50.\left(51-47\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+........+48.49.50.51-47.48.49.50\)
\(=48.49.50.51\)
\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)
XO
23 tháng 7 2019
Ta có C = 1 x 2 x 3 + 2 x 3 x 4 + ... + 48 x 49 x 50
=> 4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + .... + 48 x 49 x 50 x 4
4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 - 1)+ ... + 48 x 49 x 50 x (51 - 47)
4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + .... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
4C = 48 x 49 x 50 x 51
4C = 5997600
C = 5997600 : 4
C = 1499400
Vậy C = 1499400
C
0
\(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.4+...+8.9.10.4\)
\(4C=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)
\(4C=1.2.3.4+2.3.4.5+....+8.9.10.11\)
\(\Rightarrow C=\frac{8.9.10.11}{4}=1980\)
Ta có : C = 1 x 2 x 3 + 2 x 3 x 4 +...........+ 8 x 9 x 10
=> 4C = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ..... + 8.9.10.11
=> 4C = 8.9.10.11
=> C = \(\frac{8.9.10.11}{4}=1980\)