已知数列an满足a1+2a2+22a3+…+2n-1an=n2(n∈N*).
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已知数列an满足a1+2a2+22a3+…+2n-1an=
n |
2 |
(Ⅰ)n=1时,a1=
1
2
∵a1+2a2+22a3+…+2n-1an=
n
2…..(1)
∴n≥2时,a1+2a2+22a3+…+2n-2an=
n−1
2….(2)
(1)-(2)得2n−1an=
1
2即an=
1
2n
又a1=
1
2也适合上式,∴an=
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
=
2(1−2n)
1−2−n•2n+1=2n+1−2−n•2n+1
∴Sn=(n−1)•2n+1+2
1
2
∵a1+2a2+22a3+…+2n-1an=
n
2…..(1)
∴n≥2时,a1+2a2+22a3+…+2n-2an=
n−1
2….(2)
(1)-(2)得2n−1an=
1
2即an=
1
2n
又a1=
1
2也适合上式,∴an=
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
=
2(1−2n)
1−2−n•2n+1=2n+1−2−n•2n+1
∴Sn=(n−1)•2n+1+2
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