1+1+1+1+1+1+2+2+2+2+2+2+3+3+3+3+3+3+4x6+5x6+6x6+7x6+8x6+9x6+10x6+11x6+100x6000000000600000
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C=1/(2x4)+1/(4x6)+...+1/(18x20)
2C=2/(2x4)+2/(4x6)+...+2/(18x20)
2C=1/2-1/4+1/4-1/6+....-1/20
2C= 1/2- 1/20
2C= 9/20
C= 9/20 x 1/2
C= 9/40
- Quên k auto súc vặc
\(\frac{4}{9}\cdot\frac{11}{15}+\frac{4}{9}\cdot\frac{4}{15}\)
\(=\frac{4}{9}\left(\frac{11}{15}+\frac{4}{15}\right)\)
\(=\frac{4}{9}\cdot1=\frac{4}{9}\)
\(\left(x-\frac{2}{3}\right).\frac{1}{6}=\frac{1}{8}\)
\(\left(x-\frac{2}{3}\right)=\frac{3}{4}\)
\(x=\frac{17}{12}\)
B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004
1 x 2 x 3 x 4 x .............x 2002 x 2003
2 x 3 x 4 x .............x 2003 x 2004
1
2004
B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\frac{99}{200}\)
B = \(\frac{891}{400}\)
D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49
3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3
3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )
3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49
3D = 48 x 49 x 50
D = ( 48 x 49 x 50 ) : 3
D = 39200
E = 12 + 22 + 32 + ... + 482
E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48
E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )
E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49
E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )
Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225
thay vào ta được :
E = 39200 - 1225
E = 37975
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
`(2/(2xx4)+2/(4xx6)+2/(6xx8)+2/(8xx10))xxy=1/3`
`=>(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10)xxy=1/3`
`=>(1/2-1/10)xxy=1/3`
`=>(5/10-1/10)xxy=1/3`
`=>4/10xxy=1/3`
`=>2/5xxy=1/3`
`=>y=1/3:2/5`
`=>y=1/3xx5/2`
`=>y=5/6`
100600000000069
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