问题
解答题
已知数列an满足a1+2a2+22a3+…+2n-1an=
(Ⅰ)求数列{an}的通项; (Ⅱ)若bn=
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答案
(Ⅰ)n=1时,a1=1 2
∵a1+2a2+2a3…+2n-1an=
…..(1)n 2
∴n≥2时,a1+2a2+2a3…+2n-2an-1=
….(2)n-1 2
(1)-(2)得2n-1an=
即an=1 2 1 2n
又a1=
也适合上式,∴an=1 2 1 2n
(Ⅱ)bn=n•2n,∴Sn=1•2+2•22+3•23+…+n•2n(3)
2Sn=1•22+2•23+…+(n-1)•2n+n•2n+1(4)
(3)-(4)可得-Sn=1•2+1•22+1•23+…+1•2n-n•2n+1
=
-n•2n+1=2n+1-2-n•2n+12(1-2n) 1-2
∴Sn=(n-1)•2n+1+2