Question

(理)設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n,滿足(4-p)S_n+3pa_n=2p+4,其中p為常數(shù),且p＜-2,n∈N^*.

(1)求證:數(shù)列{a_n}是等比數(shù)列,并求出數(shù)列{a_n}的通項(xiàng)公式;

(2)若數(shù)列{a_n}的公比q=f(p),數(shù)列{b_n}滿足b₁=a₁,b_n=f(b_n-1)(n≥2),求數(shù)列{b_n}的通項(xiàng)公式;

(3)在(2)的條件下,若(b_nlga_n)=lg(p),求實(shí)數(shù)p的值.

(文)設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n,滿足(4-p)S_n+3pa_n=2p+4,其中p為常數(shù),且p＜-2,n∈N^*.

(1)求a₁并證明數(shù)列{a_n}是等比數(shù)列;

(2)若p=-4,求a₄的值;

(3)若數(shù)列{a_n}的公比q=f(p),數(shù)列{b_n}滿足b₁=a₁,b_n=f(b_n-1)(n≥2),求數(shù)列{b_n}的通項(xiàng)公式.

Answer 1

答案：(理)(1)證明:當(dāng)n=1時(shí),由(4-p)a₁+3pa₁=2p+4,得(2p+4)a₁=2p+4.∵p＜-2,2p+4≠0,∴a₁=1.

又由(4-p)S_n+3pa_n=2p+4,(4-p)S_n-1+3pa_n-1=2p+4(n≥2).

兩式相減得(4-p)a_n+3p(a_n-a_n-1)=0,(4+2p)a_n=3pa_n-1.故(n≥2).

∴數(shù)列{a_n}是以1為首項(xiàng)、為公比的等比數(shù)列.

∴a_n=.

(2)解:f(p)=,b₁=a₁=1.∵b_n=f(b_n-1)=(n≥2),

∴,即(n≥2).

故數(shù)列{}是以=1為首項(xiàng)、為公差的等差數(shù)列.

∴=1+(n-1)=.∴b_n=(n≥1).

(3)解：由(b_nlga_n)=lg(),

又b_nlga_n=lg()^n-1=,

∴(b_nlga_n)=2lg()=lg().∴()²=.

化簡(jiǎn)為p²+5p+4=0,解得p=-1與p=-4.由題意知p＜-2,故舍去-1,得p=-4.

(文)解：(1)當(dāng)n=1時(shí),由(4-p)a₁+3pa₁=2p+4,得(2p+4)a₁=2p+4.

∵p＜-2,2p+4≠0,∴a₁=1.

又由(4-p)S_n+3pa_n=2p+4,(4-p)S_n-1+3pa_n-1=2p+4(n≥2),兩式相減得(4-p)a_n+3p(a_n-a_n-1)=0.

(4+2p)a_n=3pa_n-1,故=(n≥2).

∴數(shù)列{a_n}是以1為首項(xiàng)、為公比的等比數(shù)列.

(2)當(dāng)p=-4時(shí),公比q==3.

∴a₄=a₁q³=27.9分

(3)f(p)=,b₁=a₁=1,∵b_n=f(b_n-1)=·(n≥2),

∴,即=(n≥2).故數(shù)列{}是以=1為首項(xiàng)、為公差的等差數(shù)列.

∴=1+(n-1)=.∴b_n=(n≥1).