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{a_n}的前n項和為S_n,點(n,S_n)(n∈N^*)均在函數(shù)y=f(x)的圖象上.
(1)求數(shù)列{a_n}的通項公式.
(2)若c_n=(a_n+2),2b₁+2²b₂+2³b₃+…+2ⁿb_n=c_n,求數(shù)列{b_n}的通項公式.

Answer 1

(1) a_n=6n-5   (2) b_n=

【思路點撥】(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S_n=3n²-2n.由S_n可求a_n.
(2)根據(jù)a_n求c_n,求出c_n代入2b₁+2²b₂+2³b₃+…+2ⁿb_n=c_n中可求出b_n,注意n=1與n≥2的討論.
解：(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(n,S_n)(n∈N^*)均在函數(shù)y=f(x)的圖象上,
則S_n=3n²-2n,當(dāng)n=1時,a₁=S₁=1;
當(dāng)n≥2時,a_n=S_n-S_n-1=6n-5,
故數(shù)列{a_n}的通項公式:a_n=6n-5.
(2)由(1)得,c_n=(a_n+2)=2n-1,
2b₁+2²b₂+2³b₃+…+2ⁿb_n=2n-1,
當(dāng)n=1時,b₁=,
當(dāng)n≥2時,2b₁+2²b₂+2³b₃+…+2^n-1b_n-1+2ⁿb_n
=2n-1,
2b₁+2²b₂+2³b₃+…+2^n-1b_n-1=2(n-1)-1,
兩式相減得:b_n==2^1-n,
故數(shù)列{b_n}的通項公式:b_n=