Question

观察式子:1/(1*3)=1/2(1-1/3) .1/(3*5)=1/2*(1/3-1/5).1/(5*7)=1/2*(1/5-1/7)

Answer 1

解:由这些式子的共同规律是，n为正整数时：1/[n(n+2)]=1/2*[1/n-1/(n+2)].
(1)计算:1/(1*3)+1/(3*5)+1/(5*7)+1/(7*9)+…+1/(97*99)
=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+…+1/2*(1/97-1/99)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+…+1/97-1/99)
=1/2*(1-1/99)
=1/2*(98/99)
=49/99
(2)计算:1/(1*3)+1/(3*5)+1/(5*7)+…+1/[n(n+2)]
=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+…+1/2*[1/n-1/(n+2)]
=1/2*[1-1/3+1/3-1/5+1/5-1/7+…+1/n-1/(n+2)]
=1/2*[1-1/(n+2)]
=1/2*[(n+1)/(n+2)]
=(n+1)/(2n+4)

Answer 2

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