Question

数列｛an｝满足a1=1且an=2a(n-1)+2^n(n≮2且n∈N*) 设前N项和为Sn,求证Sn/2^n>2n-3

Answer 1

an=2a(n-1)+2^n
二边同除以2^n:
an/2^n=2a(n-1)/2^n+1=a(n-1)/2^(n-1)+1
即an/2^n-a(n-1)/2^(n-1)=1
故数列{an/2^n}是一个等差数列,首项a1/2=1/2,d=1
故有an/2^n=1/2+(n-1)*1=n-1/2
an=(n-1/2)*2^n
Sn=1/2*2+3/2*2^2+5/2*2^3+....+(n-1/2)*2^n
2Sn=1/2*2^2+3/2*2^3+.....+(n-1/2)*2^(n+1)
Sn-2Sn=1/2*2+(2^2+2^3+...+2^n)-(n-1/2)2^(n+1)
-Sn=1+4*(2^(n-1)-1)/(2-1)-(n-1/2)2^n*2
Sn=-1+4-2*2^n+2n*2^n-2^n=3+(2n-3)*2^n
故有Sn/2^n=3/2^n+(2n-3)>2n-3.
得证.