⑴an=2a(n-1)+2^n
两边同时除以2^n,得
an/(2^n)=a(n-1)/[2^(n-1)]+1
即[an/(2^n)]-[a(n-1)/[2^(n-1)]]=1
∴数列{an/(2^n)}是首项a1/2=1/2,公差d=1的等差数列
∴an/2^n=1/2+(n-1)×1=n-1/2
则an=(n-1/2)×(2^n)
⑵Sn用错位相减法求即可。
Sn=1/2×2+3/2×2^2+5/2×2^3+……+(n-1/2)×(2^n) ①
则2Sn=1/2×2^2+3/2×2^3+5/2×2^4+……+(n-1/2)×(2^(n+1)) ②
①-②,得:
-Sn=1+2^2+2^3+……+2^n-(n-1/2)×(2^(n+1))
=2+2^2+2^3+……+2^n-(n-1/2)×(2^(n+1)) -1
=[2(1-2^n)]/(1-2)-(n-1/2)×(2^(n+1)) -1
=(3-2n)×(2^n)-3
∴Sn=(2n-3)×(2^n)+3>(2n-3)×(2^n)
即Sn/(2^n)>2n-3
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