Question

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

Answer 1

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