等差数列{an}中,a2=4,其前n项和Sn满足Sn＝n2+λn(λ∈R)

Question

等差数列{an}中,a2=4,其前n项和Sn满足Sn＝n2+λn(λ∈R)

Answer 1

(1)a1=S1=1+λ
S2=a1+a2=1+λ+4=4+2λ
解得λ=1
a1=1+λ=1+1=2
Sn=n²+n
Sn-1=(n-1)²+(n-1)
an=Sn-Sn-1=n²+n-(n-1)²-(n-1)=2n
n=1时,a1=2,同样满足.
数列{an}的通项公式为an=2n.
(2)Sn=n²+n=n(n+1)
1/Sn+bn=1/n-1/(n+1)+bn
1/S1+b1=1/2+b1=1
b1=1/2
1/Sn+bn=1×2^(n-1)=2^(n-1)
bn=2^(n-1)-1/Sn=2^(n-1)-[1/n-1/(n+1)]
Tn=b1+b2+...+bn
=2^0+2^1+...+2^(n-1)-[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=(2^n -1)/(2-1)-[1-1/(n+1)]
=2^n -1 -n/(n+1)