A= 1x2x3+ 2x3x4+ 3x4x5+........+ 9x10x11
ai giải được mình tích
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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 \(A=\dfrac{-4949}{19800}\)
Câu b:
B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100
B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)
B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)
B = 1/2.(1/2 - 1/9900)
B = 1/2.4949/9900
B = 4949/19800
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
A = 1.3 + 3.5 + ..+ 99.101
6A = 1.3.6 + 3.5.6 + 5.7.6 + ...+99.101.6
1.3.6 = 1.3.(5 + 1) = 1.3.5 + 1.3.1
3.5.6 = 3.5.(7 - 1) = 3.5.7 - 1.3.5
5.7.6 = 5.7.(9- 3) = 5.7.9 - 3.5.7
.........................................................
99.101.6 = 99.101.(103 - 97) = 99.101.103-97.99.101
Cộng vế với vế ta có:
6A = 1.3.1 + 99.101.103
6A = 3 + 9999.103
6A = 3 + 1029897
6A = 1029900
A = 1029900 : 6
A = 171650
Câu b:
B = 1.2.3 + 2.3.4 + 3.4.5 + ...+99.100.101
4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +..+ 99.100.101.4
1.2.3.4 = 1.2.3.4
2.3.4.4 = 2.3.4.(5-1) = 2.3.4.5 - 1.2.3.4
3.4.5.4 = 3.4.5.(6 - 2) = 3.4.5.6 - 2.3.4.5
..............................................................................
99.100.101.4 = 99.100.101.(102 - 98) = 99.100.101.102 - 98.99.100.101
Cộng vế với vế ta có:
4B = 99.100.101.102
B = 99.100.101.102 : 4
B = 25497450
\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=190/380-1/380
=189/380
Gọi biểu thức trên là S. Ta có :
\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)
Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta sẽ có :
\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)
\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)