数列an的前n项和为sn满足2sn=n(an+2)且a2,a5,a9成等比数列

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数列an的前n项和为sn满足2sn=n(an+2)且a2,a5,a9成等比数列
1求数列的通项公式
2求和：a1*a2-a2*a3-a3*a4.a2n*a2n+1

1.
2Sn=n(an+2)
2S(n+1)=(n+1)[a(n+1)+2]
2a(n+1)=(n+1)a(n+1)-nan+2
(n-1)a(n+1)-nan+2=0
当n=1时,得a1=2
a(n+1)/n-an/(n-1)+2/[n(n-1)]=0
设bn=an/(n-1),b2=a2
b(n+1)-bn=-2/[n(n-1)]=2/n-2/(n-1)
bn-b(n-1)=2/(n-1)-2/(n-2)
b(n-1)-b(n-2)=2/(n-2)-2/(n-3)
.
b4-b3=2/3-2/2
b3-b2=2/2-2/1
两边相加：
bn-b2=2/(n-1)-2/1
bn=2/(n-1)-2+a2
an/(n-1)=2/(n-1)-2+a2
an=(n-1)a2-2n+4
a5=4a2-2*5+4=4a2-6
(a5)^2=(4a2-6)^2=16(a2)^2-48a2+36
a9=8a2-2*9+4=8a2-14
a2*a9=a2(8a2-14)=8(a2)^2-14a2
16(a2)^2-48a2+36=8(a2)^2-14a2
8(a2)^2-34a2+36=0
a2=2或9/4
所以an=(n-1)2-2n+4=2或an=(n-1)9/4-2n+4=（n+7)/4
即an=2或an=（n+7)/4.
2.
(1)
当an=2时,
sn=2^2-2^2-……-2^2,(2n个)
=-（2^2+……+2^2）,(2n-2个)
=-（2n-2)2^2
=8-8n
(2)
当an=n/4+7/4时,
16s2n=8*9-9*10-10*11-……-（2n+6)(2n+7)-(2n+7)(2n+8)
144-16s2n=8*9+9*10+10*11+……+（2n+6)(2n+7)+(2n+7)(2n+8)
设cn=(n+7)(n+8)=n^2+15n+56
cn前2n项和T2n=2n(2n+1)(4n+1)/6+30n(2n+1)+112n
144-16s2n=2n(2n+1)(4n+1)/6+30n(2n+1)+112n
s2n=[144-2n(2n+1)(4n+1)/6-30n(2n+1)-112n]/16
=[144-n(2n+1)(4n+1)/3-30n(2n+1)-112n]/16

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