设{an}是等差数列,且首项a1>0,公差d>0求证:1/a1a2+1/a2a3+…+1/anan+1=n/a1(a1+nd)
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1/a1a2+1/a2a3+…+1/anan+1
= [ (a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1 ] /d
=[ 1/a1 - 1/a2 + 1/a2 - 1/a3 +...+ 1/an - 1/a(n+1) ] /d
=[ 1/a1 - 1/a(n+1) ] /d
=(a(n+1)-a1)/a1a(n+1)d
=nd / a1a(n+1)d
=n/a1(a1+nd)
= [ (a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1 ] /d
=[ 1/a1 - 1/a2 + 1/a2 - 1/a3 +...+ 1/an - 1/a(n+1) ] /d
=[ 1/a1 - 1/a(n+1) ] /d
=(a(n+1)-a1)/a1a(n+1)d
=nd / a1a(n+1)d
=n/a1(a1+nd)
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