Question

在数列{an}中,a1=2,a(n+1)=4an-3n+1,n为N* (1)设bn=nan-n^2,求数列{bn}的前n

Answer 1

a (n+1)=4an-3n+1

=>

a(n+1) - (n+1) = 4(an -n)

{an - n}是等比数列

an-n = 4^(n-1)*(a1-1)=4^(n-1)

=>

an=4^(n-1) + n

bn=nan-n^2=n*4^(n-1)

Sn=4^0+2*4^1+3*4^2+……n*4^(n-1)

4Sn=4^1+2*4^2+3*4^3+……+(n-1)*4^(n-1)+n*4^n

4Sn-Sn=-4^0-4^1-4^2-……-4^(n-1)+n*4^n

3Sn=n*4^n-(1-4^n)/(1-4)

3Sn=n*4^n+(1-4^n)/3= (3n-1)*4^n/3+1/3

Sn=(3n-1)*4^n/9+1/9