Question

一道简单的数列题,我是新来的已知数列{an}满足a1=1,an+₁-2an=2n-1（1）求数列{an}的通

Answer 1

(1)∵数列{a[n]}满足a[1]=1,a[n+1]-2a[n]=2n-1

∴a[n+1]=2a[n]+2n-1 【1】

用待定系数法：a[n+1]+x(n+1)+y=2(a[n]+xn+y)

将【1】代入：2a[n]+2n-1+xn+x+y=2a[n]+2xn+2y

即：(2+x)n+(x+y-1)=2xn+2y

∴2+x=2x,得：x=2

x+y-1=2y,得：y=1

∴a[n+1]+2(n+1)+1=2(a[n]+2n+1)

∵a[1]+2+1=4

∴{a[n]+2n+1}是首项为4,公比为2的等比数列

∴a[n]+2n+1=4*2^(n-1)=2^(n+1)

即：a[n]=2^(n+1)-2n-1

(2)S[n]=a[1]+a[2]+...+a[n]

=(2^2-2*1-1)+(2^3-2*2-1)+...+[2^(n+1)-2n-1]

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

=2^(n+2)-(n^2+2n+4)