Question

计算题3道，简单说一下解题思路

1.

2.

3.

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Answer 1

1/(2²-1)+1/(4²-1)+1/(6²-1)+......+1/(100²-1)
=1/[(2-1)(2+1)]+1/[(4-1)(4+1)]+1/[(6-1)(6+1)]+......+1/[(100-1)(100+1)]
=1/(1*3)+1/(3*5)+1/(5*7)+......+1/(99*101)
=[2/(1*3)+2/(3*5)+2/(5*7)+......+2/(99*101)]/2
=[(1-1/3)+(1/3-1/5)+(1/5-1/7)+......+(1/99-1/101)]/2
=(1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101)/2
=(1-1/101)/2
=(100/101)/2
=50/101

Answer 2

1. n^2-1=(n+1)(n-1),1/[n^2-1]=0.5*(1/(n-1)-1/(n+1))然后大多数项都约

2. 全化成假分数，答案1

3. 分子分母提出公因式1.1