Question

已知数列{an}的各项均为正数，其前n项和Sn=1/2(an-1)(an+2),n€N*. (1)求{an}通

已知数列{an}的各项均为正数，其前n项和Sn=1/2(an-1)(an+2),n€N*. (1)求{an}通项公式，(2)设bn=(-1)^nanan+1求{bn}的前2n项和T2n

Answer 1

(1)
Sn=(1/2)(an-1)(an+2)
n=1,
a1=(1/2)(a1-1)(a1+2)
(a1)^2-a1-2=0
(a1-2)(a1+1)=0
a1=2

for n>=2
an = Sn -S(n-1)
2an =(an-1)(an+2) - (a(n-1)-1)(a(n-1)+2)
(an)^2-an - [a(n-1)]^2- a(n-1) =0
[an + a(n-1)]. [an - a(n-1)-1]=0
an - a(n-1)-1=0
an -a1 = n-1
an = n+1
(2)
bn=(-1)^n.an.a(n+1)
= (-1)^n . (n+1)(n+2)

cn =b(2n-1) + b(2n)
=(-1)^(2n-1) . 2n(2n+1) + (-1)^(2n) . (2n+1)(2n+2)
=(2n+1)(2n+2) - 2n(2n+1)
= 2(2n+1)

T(2n)=b1+b2+...+b(2n)
= c1+c2+...+cn
= 2( 2n+4) n/2
= 2(n+2)n