Question

已知二次函數(shù)f(x)=px2+qx(p≠0),其導(dǎo)函數(shù)為f'(x)=6x-2,數(shù)列{an}的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)(n∈N*)均在函數(shù)y=f(x)的圖象上.

(1)求數(shù)列{an}的通項(xiàng)公式.

(2)若cn=(an+2),2b1+22b2+23b3+…+2nbn=cn,求數(shù)列{bn}的通項(xiàng)公式.

 

Answer 1

(1) an=6n-5 (2) bn=

【解析】【思路點(diǎn)撥】(1)根據(jù)二次函數(shù)的導(dǎo)函數(shù)為f'(x)=6x-2,可求f(x)=3x2-2x,所以Sn=3n2-2n.由Sn可求an.

(2)根據(jù)an求cn,求出cn代入2b1+22b2+23b3+…+2nbn=cn中可求出bn,注意n=1與n≥2的討論.

【解析】
(1)已知二次函數(shù)f(x)=px2+qx(p≠0),

則f'(x)=2px+q=6x-2,故p=3,q=-2,

所以f(x)=3x2-2x.

點(diǎn)(n,Sn)(n∈N*)均在函數(shù)y=f(x)的圖象上,

則Sn=3n2-2n,當(dāng)n=1時(shí),a1=S1=1;

當(dāng)n≥2時(shí),an=Sn-Sn-1=6n-5,

故數(shù)列{an}的通項(xiàng)公式:an=6n-5.

(2)由(1)得,cn=(an+2)=2n-1,

2b1+22b2+23b3+…+2nbn=2n-1,

當(dāng)n=1時(shí),b1=,

當(dāng)n≥2時(shí),2b1+22b2+23b3+…+2n-1bn-1+2nbn

=2n-1,

2b1+22b2+23b3+…+2n-1bn-1=2(n-1)-1,

兩式相減得:bn==21-n,

故數(shù)列{bn}的通項(xiàng)公式:bn=