已知数列{an}的前n项和为Sn=n2+n求数列{an}的通项公式；若bn=(1/2)an+n,求数列{bn}的前n项和Tn

Question

急求

Answer 1

解：
(1)
a1=S1=1^2+1=2
Sn=n^2+n
Sn-1=(n-1)^2+(n-1)
an=Sn-Sn-1=n^2+n-(n-1)^2-(n-1)=2n
{an}通项公式为an=2n
(2)
bn=(1/2)^an+n=(1/2)^(2n)+n=(1/4)^n+n
Tn=b1+b2+...+bn
=(1/4)^1+(1/4)^2+...+(1/4)^n+(1+2+...+n)
=(1/4)[(1-(1/4)^n]/(1-1/4)+n(n+1)/2
=1/3-(1/3)(1/4)^n+n(n+1)/2

Answer 2

1).Sn=n^2+n，递推，S(n-1)=（n-1)^2+n-1。二式相减，an=n^2-(n-1)^2+1=n^2-(n^2-2n+1)+1=2n-1+1=2n，an=2n。2).bn=（1/2)2n+n=（2/2)n+n=n+n=2n，bn=2n。Tn=Σbn=Σ2n=2Σn=2n（n+1)/2=n(n+1)。

Answer 3

n=1时 A1=2An=Sn-S（n-1）=2n，Bn=2n，则Tn=n^2+n

Answer 4

讨论得an=2n. tn=n的平方+n