Question

已知数列an中，a1=2,an+1=4an-3n+1，bn=an-n,求证数列bn为等比数列，求an前n项和

Answer 1

an+1=4an-3n+1
an+1-n-1=4an-4n
an+1-（n+1)=4(an-n)
[an+1-(n+1)]/(an-n)=4
所以an-n是等比数列
bn=an-n，所以bn是等比数列，b1=1，q=4
Sbn=(4^n-1)/3
San=Sbn-(1+2+3.....+n)
=(4^n-1)/3-(1+n)*n/2
=(4^n-1)/3-(n+n^2)/2

Answer 2

a(n+1) = 4a(n) - 3n + 1,
a(n+1) - (n+1) = 4a(n) - 4n = 4[a(n) - n],
b(n) = a(n) - n,
{b(n)}是首项为b(1)=a(1)-1=1,公比为4的等比数列。
a(n)-n=b(n)=4^(n-1),
a(n) = n + 4^(n-1), n = 1,2,..
S(n) = a(1) + a(2) + ... + a(n)
= 1 + 1 + 2 + 4 + ... + n + 4^(n-1)
= 1 + 2 + ... + n + 1 + 4 + ... + 4^(n-1)
= n(n+1)/2 + [4^n - 1]/(4-1)
= n(n+1)/2 + (4^n - 1)/3.