1/1*2+1/2*3+1/3*4++1/(X+1)(x+2)= 2011/2012
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1/1*2+1/2*3+1/3*4++1/(X+1)(x+2)= 2011/2012
注意到:
1/1*2 = (2-1)/1*2=1/1 - 1/2
1/2*3 = (3-2)/2*3=1/2 - 1/3
1/3*4 = (4-3)/3*4=1/3 - 1/4
.
1/(X+1)(x+2) = 1/(x+1) - 1/(x+2)
因此:上述各式相加,则:
左边=1/1*2+1/2*3+1/3*4+.+1/(X+1)(x+2)
右边=1/1 - 1/2+1/2 - 1/3+1/3 - 1/4+1/(x+1) - 1/(x+2)
=1-1/(x+2)
因此,原方程为:
1-1/(x+2) = 2011/2012
即:
(x+1)/(x+2) = 2011/2012
∴
2012(x+1)=2011(x+2)
2012x+2012=2011x+4022
x=2010
1/1*2 = (2-1)/1*2=1/1 - 1/2
1/2*3 = (3-2)/2*3=1/2 - 1/3
1/3*4 = (4-3)/3*4=1/3 - 1/4
.
1/(X+1)(x+2) = 1/(x+1) - 1/(x+2)
因此:上述各式相加,则:
左边=1/1*2+1/2*3+1/3*4+.+1/(X+1)(x+2)
右边=1/1 - 1/2+1/2 - 1/3+1/3 - 1/4+1/(x+1) - 1/(x+2)
=1-1/(x+2)
因此,原方程为:
1-1/(x+2) = 2011/2012
即:
(x+1)/(x+2) = 2011/2012
∴
2012(x+1)=2011(x+2)
2012x+2012=2011x+4022
x=2010
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