Question

已知以a₁为首项的数列{a_n}满足:a_n+1=

(1)当a₁=1,c=1,d=3时,求数列{a_n}的通项公式;

(2)当0＜a₁＜1,c=1,d=3时,试用a₁表示数列{a_n}前100项的和S₁₀₀;

(3)当0＜a₁＜(m是正整数),c=,正整数d≥3m时,求证:数列a₂-,a_3m+2-,a_6m+2-,a_9m+2-成等比数列当且仅当d=3m.

Answer 1

解:(1)由题意得a_n=(k∈Z⁺).

(2)当0＜a₁＜1时,

a₂=a₁+1,a₃=a₁+2,a₄=a₁+3,a₅=+1,a₆=+2,a₇=+3,…,a_3k-1=+1,a_3k=+2,a_3k+1=+3,

∴S₁₀₀=a₁+(a₂+a₃+a₄)+(a₅+a₆+a₇)+…+(a₉₈+a₉₉+a₁₀₀)

=a₁+(3a₁+6)+(a₁+6)+(+6)+…+(+6)

=a₁+a₁(3+1++…+)+6×33

=(11-)a₁+198.

(3)当d=3m时,a₂=a₁+,

∵a_3m=a₁+=a₁-+3＜3＜a₁+3=a_3m+1,

∴a_3m+2=+.

∵a_6m=-+3＜3＜+3=a_6m+1,

∴a_6m+2=+.

∵a_9m=-+3＜3＜+3=a_9m+1,

∴a_9m+2=+.

∴a₂-=a₁,a_3m+2-=,a_6m+2-=.

∴a_9m+2-=.

综上所述,当d=3m时,数列a₂-,a_3m+2-,a_6m+2,a_9m+2-是公比为的等比数列.

当d≥3m+1时,a_3m+2=∈(0,),

a_6m+2=+3∈(3,3+),a_6m+3=∈(0,),

a_9m+2=+∈(3-,3),

由于a_3m+2-＜0,a_6m+2-＞0,a_9m+2-＞0.

故数列a₂-,a_3m+2-,a_6m+2-,a_9m+2-不是等比数列.

所以,数列a₂-,a_3m+2-,a_6m+2-,a_9m+2-成等比数列,      

当且仅当d=3m.