等差数列an=5n-4,求{1/an.an+1}的前n项和.
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∵1/(an·an+1)=1/[(5n-4)(5n+1)]=(1/5)[1/(5n-4)-1/(5n+1)],
∴1/(a1·a2)+1/(a2·a3)+1/(a3·a4)+······+1/(an·an+1)
=1/(1×6)+1/(6×11)+1/(11×16)+······+1/[(5n-4)(5n+1)]
=(1/5)[1-1/6+1/6-1/11+1/11-1/16+······+1/(5n-4)-1/(5n+1)]
=(1/5)[1-1/(5n+1)]
=n/(5n+1).
∴1/(a1·a2)+1/(a2·a3)+1/(a3·a4)+······+1/(an·an+1)
=1/(1×6)+1/(6×11)+1/(11×16)+······+1/[(5n-4)(5n+1)]
=(1/5)[1-1/6+1/6-1/11+1/11-1/16+······+1/(5n-4)-1/(5n+1)]
=(1/5)[1-1/(5n+1)]
=n/(5n+1).
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