Question

已知数列{an}满足a1=1,且an=2a(n-1)+2^n(n≥2且n∈N*)

‍‍（1）{an}的通向公式
（2）{an}的前n项和为sn。求证sn/2^n＞2n-3。‍

Answer 1

(1)
an/2^n=1/2+(n-1)=n-1/2
an=n2^n-2^(n-1)

(2)
sn=[2^1-2^0]+[2*2^2-2^1]+[3*2^3-2^2]+……+[(n-2)2^(n-2)-2^(n-3)]+[(n-1)2^(n-1)-2^(n-2)]+[n2^n-2^(n-1)]
=-2^0+2^2+2*2^3+……+(n-3)2^(n-2)+(n-2)2^(n-1)+n2^n

sn=-2^0+2^2+2*2^3+……+(n-3)2^(n-2)+(n-2)2^(n-1)+n2^n
2sn=-2^1+2^3+2*2^4+……+(n-3)2^(n-1)+(n-2)2^n+n2^(n+1)
两式相减：
sn=2^0-2^1-2^2-2^3-……-2^(n-2)-2^(n-1)-2^n+n2^(n+1)
=2-2^0-2^1-2^2-2^3-……-2^(n-2)-2^(n-1)-2^n+n2^(n+1)
=2-[2^(n+1)-1]/(2-1)+n2^(n+1)
=3-2^(n+1)+n2^(n+1)
=3+(n-1)2^(n+1)
sn=3+(n-1)2^(n+1)
sn/2^n=3/2^n+2n-2
3/2^n＞0
sn/2^n＞2n-2＞2n-3
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