Question

已知数列{an}满足的前n项和为sn,且sn=(1/3)∧n+n-1,(n∈n*).(1)求数列{an}通项公式.(2)

Answer 1

an=Sn-S(n-1)=1-2*(1/3)^n

bn=2n*(1/3)^n

Tn= 2* [ 1/3 + 2*1/(3^2) + 3*1/(3^3) + ..+ n*1/(3^n) ]

3Tn=2* [ 1 + 2*1/3 3*1/(3^2) + 4*1/(3^3) + .+ n*1/(3^n-1) ]

错位相减

3Tn-Tn=2Tn=2*{1+ 1/3+1/(3^2)+1/(3^3)+.1/[3^(n-1)] - n*1/(3^n) }

Tn=1+1/3+.1/[3^(n-1)] - n*1/(3^n) = 3/2 [ 1- (1/3)^n ] - n*1/(3^n)

Tn=3/2-(2n+3)/2*(1/3)^n