Question

設(shè)S_n是數(shù)列{a_n}的前n項(xiàng)和,滿(mǎn)足S_n=(1-a_n)(a＞0且a≠1,n∈N^*).數(shù)列{b_n}滿(mǎn)足b_n=a_nlga_n(n∈N^*).

(1)求數(shù)列{a_n}的通項(xiàng)公式;

(2)若數(shù)列{b_n}中每一項(xiàng)總小于它后面的項(xiàng),求a的取值范圍.

Answer 1

解：(1)∵S_n=(1-a_n)(a＞0且a≠1),

    則S_n+1=(1-a_n+1).由S_n+1-S_n=a_n+1得a_n+1=(a_n-a_n+1),

    即a_n+1=a·a_n.

    當(dāng)n=1時(shí),a₁=(1-a₁),a₁=a,

∴數(shù)列{a_n}是以a為首項(xiàng),公比為a的等比數(shù)列.

∴數(shù)列{a_n}的通項(xiàng)公式a_n=aⁿ.

(2)依題意得b_n=naⁿlga,

    令b_k+1＞b_k,則(k+1)a^k+1lga＞ka^klga.

∵a＞0且a≠1,∴a^k＞0.

∴(k+1)alga＞klga.

①當(dāng)a＞1時(shí),由a(k+1)-k＞0得a＞.

∵0＜＜1,

∴a＞1時(shí),b_k+1＞b_k成立.

②當(dāng)0＜a＜1時(shí),lga＜0,解得a＜.

    為使不等式對(duì)任意的正整數(shù)k都成立,只需a小于的最小值.

∵=≥,解得0＜a＜.

∴a的取值范圍為{a|a＞1或0＜a＜}.