已知数列an的前n项和为Sn,a1=2,a2=1,且点（Sn,S(n+1)）在直线y=kx+2上
问题描述：
已知数列an的前n项和为Sn,a1=2,a2=1,且点（Sn,S(n+1)）在直线y=kx+2上
1.求k的值 2.求证an为等比数列 3.记Tn为数列｛Sn｝的前n项和,求T5的值
答
s1=a1=2
s2=a1+a2=2+1=3
点（S1,S2）在直线y=kx+2上
S2=S1k+2
3=2k+2
2k=1
k=1/2
y=x/2+2
S(n+1)=Sn/2+2
2S(n+1)=Sn+4
2S(n+1)-8=Sn-4
2[S(n+1)-4]=Sn-4
[S(n+1)-4]/[Sn-4]=1/2
所以Sn-4是以1/2公比的等数列
Sn-4=(S1-4)*(1/2)^(n-1)
Sn-4=(-2)*(1/2)^(n-1)
Sn-4=-(1/2)^(n-2)
Sn=4-(1/2)^(n-2)
S(n-1)=4-(1/2)^(n-3)
an=Sn-S(n-1)
=4-(1/2)^(n-2)-[4-(1/2)^(n-3)]
=(1/2)^(n-3)-(1/2)^(n-2)
=(1/2)^(n-3)-(1/2)*(1/2)^(n-3)
=(1/2)^(n-3)(1-1/2)
=(1/2)^(n-3)*1/2
=(1/2)^(n-2)
所以an是以1/2公比的等数列
T5=a1(1-q^5)/(1-q)
=2*[1-(1/2)^5]/(1-1/2)
=4*[1-(1/2)^5]
=4-4*1/32
=4-1/8
=31/8