Question

設(shè){a_n}是等差數(shù)列,{b_n}是各項(xiàng)都為正數(shù)的等比數(shù)列,且a₁=b₁=1,a₃+b₅=21,a₅+b₃=13.
(1)求{a_n},{b_n}的通項(xiàng)公式.
(2)求數(shù)列{}的前n項(xiàng)和S_n.

Answer 1

(1) a_n=2n-1   b_n=2^n-   (2) S_n=6-

(1)設(shè){a_n}的公差為d,{b_n}的公比為q,則依題意有q>0且解得
所以a_n="1+(n-1)" d=2n-1,b_n=q^n-1=2^n-1.
(2)=,
S_n=1+++…++,　①
2S_n=2+3++…++.　②
②-①,得S_n=2+2+++…+-
=2+2×(1+++…+)-,
=2+2×-=6-.