Question

详细解答一道数学题-1/1*2-1/2*3-1/3*4...-1/2012*2013 顺便说说这种题型怎么解

Answer 1

这类题用裂项法求解,裂项后相加时好多项会相互抵消,从而可以简化运算.
解:-1/1*2-1/2*3-1/3*4...-1/2012*2013
=-[1/(1*2)+1/(2*3)+1/(3*4)+…+1/(2012*2013)]
=-(1-1/2+1/2-1/3+1/3-1/4+…+1/2012-1/2013)
=-(1-1/2013)
=- 2012/2013.
◆比如，在算1/2-1/3时，要先通分，即：1/2-1/3=3/(2*3)-2/(2*3)=1/(2*3).
倒过来就是：1/(2*3)=1/2-1/3.
【当分母中两因数的差等于分子时,可拆为这两个因数倒数的差.】
再如:(1) 2/(1*3)+2/(3*5)+2/(5*7)+2/(7*9)
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+(1/7-1/9)
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
=1-1/9
=8/9.
(2) 1/(1*3)+1/(3*5)+1/(5*7)+1/(7*9) ----当分母中两个因数的差与分子不相等时要适当变形
=(1/2)*[2/(1*3)+2/(3*5)+2/(5*7)+2/(7*9)]
=(1/2)*(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)
=(1/2)*(1-1/9)
=4/9

Answer 2

1/(1×2)=1/1－1/2
1/(2×3)=1/2－1/3
1/(3×4)=1/3－1/4
1/(4×5)=1/4－1/5
1/(5×6)=1/5－1/6
1/(6×7)=1/6－1/7
…………
1/(2012×2013)=1/2012－1/2013
当你全部相加的时候，发现什么了？？

Answer 3

规律1/n(n+1)=1/n-1/(n+1)-1/1*2-1/2*3-1/3*4...-1/2012*2013

=-(1/1×2+1/2×3+1/3×4+…+1/2012×2013)
=-(1-1/2+1/2-1/3+1/3-1/4+...++1/2011-1/2012+1/2012-1/2013）
=-（1-1/2013）
=-2012/2013

Answer 4

-1/1*2-1/2*3-1/3*4...-1/2012*2013

=-1+1/2-1/2+1/3-1/3+1/4-……-1/2012+1/2013
=-1+1/2013
=-2012/2013

Answer 5

1/n*(n+1)=1/n-1/(n+1)每个都这样拆