等差数列，a1=1,d=4,求[1/(a1a2)]+[1/(a2a3)]+......+[1/(an-1*an)]=?

来源：百度知道编辑：UC知道时间：2024/05/09 05:29:05

an=1+4(n-1)=4n-3

[1/(a1*a2)]+[1/(a2*a3)]+......+[1/(an-1*an)]
=1/1*5+1/5*9+……+1/(4n-7)(4n-3)
=1/4(1-1/5)+1/4(1/5-1/9)+……+1/4[1/(4n-7)-1/(4n-3)]
=1/4{(1-1/5+1/5-1/9+……+1/(4n-7)-1/(4n-3)]
=1/4[1-1/(4n-3)]
=(n-1)/(4n-3)

an=a1+4(n-1)=4n-3
an-1=4n-7
1/(an-1*an)=[1/a(n-1)-1/an]/4
1/(a1*a2)]+[1/(a2*a3)]+......+[1/(an-1*an)
=[1/a1-1/a2+1/a2-1/a3+-------+1/a(n-1)-1/an]/4
=[1-1/(4n-3)]/4
=(n-1)/(4n-3)

an=1+4(n-1)
[1/(a1*a2)]+[1/(a2*a3)]+......+[1/(an-1*an)]=1/4*(1/a1-1/a2+1/a2-1/a3+......1/an-1-1/an)=1/4(1/an-1/a1)=1/4(1/1-1/(4n-3))=(n-1)/(4n-3)

仍然是个等比数列，b1=1/4,D=1/16

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等差数列，a1=1,d=4,求[1/(a1*a2)]+[1/(a2*a3)]+......+[1/(an-1*an)]=?

等差数列，a1=1,d=4,求[1/(a1a2)]+[1/(a2a3)]+......+[1/(an-1*an)]=?