Question

a1=1 an=2an-1+2n 求a2.a3 求证an/2n是等差数列 求数列an的前n项之和Sn

Answer 1

an =2a[n-1]+2^n ①

a1=1

a2=2a1+2^2=2+4=6

a3=2a2+2^3=20

an=2a(n-1)+2^n

二边同除以2^n

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

即an/2^n-a(n-1)/2^(n-1)=1

所以,{an/2^n}是一个首项是a1/2=1/2,公差是1的等差数列.

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

即an=2^n*(n-1/2)=n*2^n-2^(n-1)

Sn=1*2+2*2^2+...+n*2^n-(1+2+2^2+...+2^(n-1))

设Tn=1*2+2*2^2+...+n*2^n

2Tn=1*2^2+2*2^3+...n*2^(n+1)

Tn-2Tn=-Tn=2+2^2+2^3+2^n-n*2^(n+1)=2(2^n-1)/(2-1)-2n*2^n=2*2^n-2-2n*2^n=(2-2n)*2^n-2

即Tn=2-(2-2n)*2^n

所以,Sn=2-(2-2n)*2^n-1*(2^n-1)/(2-1)

=3-(1-2n)*2^n