已知数列{an}的前n项和为Sn,a1=1,a2=2,且点(Sn,Sn+1)在直线y=kx+1上
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已知数列{an}的前n项和为Sn,a1=1,a2=2,且点(Sn,Sn+1)在直线y=kx+1上
(Ⅰ)求k的值;
(Ⅱ)求证:{an}是等比数列;
(Ⅲ)记Tn为数列{Sn}的前n项和,求T10的值.
(Ⅰ)求k的值;
(Ⅱ)求证:{an}是等比数列;
(Ⅲ)记Tn为数列{Sn}的前n项和,求T10的值.
![已知数列{an}的前n项和为Sn,a1=1,a2=2,且点(Sn,Sn+1)在直线y=kx+1上](/uploads/image/z/15928019-35-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%EF%BC%8Ca1%3D1%EF%BC%8Ca2%3D2%EF%BC%8C%E4%B8%94%E7%82%B9%EF%BC%88Sn%EF%BC%8CSn%2B1%EF%BC%89%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dkx%2B1%E4%B8%8A)
(1)Sn+1=k•Sn+1,令n=1有,S2=k•S1+1,∴a1+a2=k•a1+1.代入a1=1,a2=2有k=2.
(2)∵Sn+1=2Sn+1,∴Sn=2Sn-1+1(n≥2).
两式相减有,an+1=2an,即,
an+1
an=2.且
a2
a1=2符合.
∴{an}为公比为2的等比数列.
(3)Sn=
1−2n
1−2=2n−1
∴T10=(2+22+23++210)−10=
2(1−210)
1−2−10=2036.
(2)∵Sn+1=2Sn+1,∴Sn=2Sn-1+1(n≥2).
两式相减有,an+1=2an,即,
an+1
an=2.且
a2
a1=2符合.
∴{an}为公比为2的等比数列.
(3)Sn=
1−2n
1−2=2n−1
∴T10=(2+22+23++210)−10=
2(1−210)
1−2−10=2036.
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