求和:1+3a+5a^2+…+(2n+1)a^n
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求和:1+3a+5a^2+…+(2n+1)a^n求和:1+3a+5a^2+…+(2n+1)a^n求和:1+3a+5a^2+…+(2n+1)a^n设S=1+3a+5a^2+…+(2n+1)a^n则aS=
求和:1+3a+5a^2+…+(2n+1)a^n
求和:1+3a+5a^2+…+(2n+1)a^n
求和:1+3a+5a^2+…+(2n+1)a^n
设S=1+3a+5a^2+…+(2n+1)a^n
则aS=a+3a^2+5a^3+…+(2n+1)a^(n+1)
两式相减下就容易了~
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