Question

设Sn是正项数列an的前n项和,且Sn=1/4an^2十1/2an一3/4

Answer 1

解：
1、当n=1时，由Sn=(1/4)an²+(1/2)an-3/4及a1=S1
得a1=(1/4)a1²+(1/2)a1-3/4
解得a1=3
当n≥2时，有an=Sn-S(n-1)
即an=[(1/4)an²+(1/2)an-3/4]-[(1/4)a²(n-1)+(1/2)a(n-1)-3/4]
化简得[an +a(n-1)].[an - a(n-1) -2 ]=0
因为an是正项数列，所以an>0
于是an+a(n-1)>0
所以只有an-a(n-1)-2=0
即an-a(n-1)=2
所以数列{an}是以a1=3为首项，2为公差的等差数列
所以an=3+2(n-1)=2n+1
当n=1时，a1=3适合an=2n+1
所以数列{an}的通项公式是an=2n+1
2、设Cn=anbn=(2n+1)2^n=n*2^(n+1)+2^n
设数列{n*2^(n+1)}前n项和为Pn ， 设数列{2^n}前n项和为Sn
于是有Tn=Pn+Sn
Pn=2²+2*2³+3*2^4+.....+n*2^(n+1)
2Pn=2³+2*2^4+3*2^5+....+(n-1)*2^(n+1)+n*2^(n+2)
上两式错项相减得-Pn=2²+2³+2^4+2^(n+1)-n*2^(n+2)
-Pn=4(2^n-1)-n*2^(n+2)
Pn=(n-1)*2^(n+2)+4
Sn=2+2²+2³+....+2^n=2(2^n-1)=2^(n+1)-2
Tn=Pn+Sn=(n-1)*2^(n+2)+4+2^(n+1)-2=(2n-1)*2^(n+1)+2

Answer 2

(1)
Sn=(1/4)(an)^2+(1/2)an-3/4
n=1,
(a1)^2-2an-3=0
(a1-3)(a1+1)=0
a1=3

an = Sn -S(n-1)
= (1/4)[(an)^2 - [a(n-1)]^2 ]+(1/2)[ an - a(n-1) ]
(an)^2 - [a(n-1)]^2 -2an-2a(n-1)=0
[an +a(n-1)].[an - a(n-1) -2 ]=0
an - a(n-1) -2=0
an - a1=2(n-1)
an= 2n+1
(2)
let
S = 1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) - ( 2+2^2+...+2^n)
= n.2^(n+1) - 2( 2^n-1)
bn=2^n
cn = an.bn
= (2n+1).2^n
=2(n.2^n) + (2^n)
Tn =c1+c2+...+cn
= 2S+ 2(2^n-1)
=2n.2^(n+1) - 4( 2^n-1) +2(2^n-1)
=2n.2^(n+1) - 2( 2^n-1)
= 2+ (4n-2).2^n