Question

设{a_n}是等差数列,{b_n}是各项都为正数的等比数列,且a₁=b₁=1,a₃+b₅=21,a₅+b₃=13.

(1)求{a_n},{b_n}的通项公式.

(2)求数列{}的前n项和S_n.

Answer

(1) an=2n-1 bn=2n- (2) Sn=6-

【解析】(1)设{an}的公差为d,{bn}的公比为q,则依题意有q>0且解得

所以an=1+(n-1) d=2n-1,bn=qn-1=2n-1.

(2)=,

Sn=1+++…++, ①

2Sn=2+3++…++. ②

②-①,得Sn=2+2+++…+-

=2+2×(1+++…+)-,

=2+2×-=6-.