题目内容
(08年北师大附中月考)已知函数f (x ) = kx + b(k≠0),f (4) = 10,又f (1),f (2),f (6)成等比数列.
(1)求函数f (x )的解析式;
(2)设an = 2f (n ) + 2n,求数列{an}的前n项和Sn.
解析:(1)∵ f (4) = 10,又f (1),f (2),f (6)成等比数列,
∴ ,解得:k = 3,b =-2,
∴ f (x ) = 3x-2.
(2)an =2f (n ) + 2n = 23n-2 + 2n = ×8n + 2n.
∴ Sn =+ n (n + 1) =
(8n-1) + n2 + n.
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