已知数列{an}满足a1=1,且an=2an-1+2n(n≥2,且n∈N*)
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已知数列{an}满足a1=1,且an=2an-1+2n(n≥2,且n∈N*)
(1)求证:数列{
(1)求证:数列{
a
(1)证明:∵an=2an-1+2n(≥2,且n∈N*)
∴ an 2n= an−1 2n−1+1 ∴ an 2n− an−1 2n−1=1 ∴数列{ an 2n}是以 1 2为首项,1为公差的等差数列; (2)由(1)得 an 2n= 1 2+(n−1)•1=n− 1 2 ∴an=(n− 1 2)•2n; (3)∵Sn= 1 2•21+ 3 2•22+…+(n− 1 2)•2n ∴2Sn= 1 2•22+ 3 2•23+…+(n− 1 2)•2n+1 两式相减可得-Sn=1+22+23+…+2n-(n− 1 2)•2n+1=(3-2n)•2n-3 ∴Sn=(2n-3)•2n+3>(2n-3)•2n ∴ Sn 2n>2n−3.
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