Question

设数列{a_n}的前n项和为S_n,已知a₁=1,且a_n+2S_nS_n-1=0(n≥2),

(1)求数列{S_n}的通项公式;

(2)设S_n=,b_n=f()+1.

记P_n=S₁S₂+S₂S₃+…+S_nS_n+1,T_n=b₁b₂+b₂b₃+…+b_nb_n+1,试求T_n,并证明P_n＜.

Answer 1

(1)解:∵a_n+2S_nS_n-1=0(n≥2),∴S_n-S_n-1+2S_nS_n-1=0.

∴=2.又∵a₁=1,∴S_n=(n∈N₊).

(2)证明:∵S_n=,∴f(n)=2n-1.∴b_n=2()-1+1=()^n-1.

T_n=()⁰·()¹+()¹·()²+…+()^n-1·()ⁿ

=()¹+()³+()⁵+…+()^2n-1=［1-()ⁿ］.

∴P_n=++…+=(1)＜.