若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an=

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若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an=

若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an=
若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an=

若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an=
因为a1+3a2+3^2a3+···+3^(n-1)an=n/2,所以a1+3a2+3^2a3+···+3^(n-2)a(n-1)=(n-1)/2
两式相减即有3^(n-1)an=n/2-(n-1)/2,化简得:an=1/[2*3^(n-1)] (n∈N*)

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