Question

已知数列{an}满足a1=1,a(n+1)=2an+n+1，设bn=an+n+2

证明数列{bn}是等比数列,2.设数列{an}的前n项和为Sn，求an和Sn

Answer 1

a(n+1)=2an+n+1
an=2a(n-1)+n
a(n+1)-an=2(an-a(n-1))+1
a(n+1)-an+1=2(an-a(n-1))+2
a(n+1)-an+1=2(an-a(n-1)+1)
a(n+1)-an+1/an-a(n-1)+1=2
又bn=an+n+2=a(n+1)-an+1
bn-1=a(n-1)+n+1=an-a(n-1)+1
bn/bn-1=a(n+1)-an+1/an-a(n-1)+1=2
b1=a1+1+2=5
bn为首项为5，公比为2的比数列
bn=5*2^(n-1)
an+n+2=5*2^(n-1)
an=5*2^(n-1)-(n+2)
sn转化为求一个等比数列一个等差数列的和
sn=5[2^0+2^1+……2^(n-1)]-[3+4+……+(n+2)]
=5*2^n-n*(n+1)/2