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∵Sn,Tn分别是等差数列{an},{bn}的前n项和, 且 Sn Tn= 2n+1 4n-2,(n∈N+), ∴ a10 b3+b18+ a11 b6+b15= a10+a11 b10+b11 = S20 T20= 40+1 80-2= 41 78. 故答案为: 41 78.
两等差数列{an}、{bn}的前n项和分别为Sn、Tn,且SnTn=5n+32n+7,则a5b5的值是( )
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