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
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![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](/uploads/image/z/13322309-5-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D2%2Ca2%3D1%2C%E4%B8%94%E7%82%B9%EF%BC%88Sn%2CS%28n%2B1%29%EF%BC%89%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dkx%2B2%E4%B8%8A1.%E6%B1%82k%E7%9A%84%E5%80%BC+2.%E6%B1%82%E8%AF%81an%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97+3.%E8%AE%B0Tn%E4%B8%BA%E6%95%B0%E5%88%97%EF%BD%9BSn%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82T5%E7%9A%84%E5%80%BC)
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
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
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
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