∴a2n+2=2×2 n1,∴a2n=2 n-2.

又a2n+a2n1= a2n2a2n+1=3a2n+1,∴数列{an}的前2007项的和为

a1+( a2+ a3)+ ( a4+ a5)+ ( a6+ a7)+ …+ ( a2006+ a2007

= a1+(3a2+1)+ (3a4+1)+ (3a6+1)+ …+ (3a2006+1)

= 1+(3×2-5)+ (3×22-5)+ (3×23-5)+ …+ (3×21003-5)

= 1+(3×2-5)+ (3×22-5)+ (3×23-5)+ …+ (3×21003-5)

= 3×(2+22+23+…+21003+1-5×1003

=6×(21003-1)+1-5×1003=6×21003- 5020 ,故选D.

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