Question

設(shè){a_n}是公比為正數(shù)的等比數(shù)列,a₁=2,a₃=a₂+4,
(1)求{a_n}的通項(xiàng)公式;
(2)設(shè){b_n}是首項(xiàng)為1,公差為2的等差數(shù)列,求數(shù)列{a_n+b_n}的前n項(xiàng)和S_n.

Answer 1

(1) a_n=2·2^n-1=2ⁿ(n∈N^*)    (2) S_n=2ⁿ⁺¹+n²-2

解:(1)設(shè)q為等比數(shù)列{a_n}的公比,
則由a₁=2,a₃=a₂+4,
得2q²=2q+4,即q²-q-2=0,
解得q=2或q=-1(舍去),因此q=2.
所以{a_n}的通項(xiàng)公式為a_n=2·2^n-1=2ⁿ(n∈N^*).
(2)∵{b_n}是等差數(shù)列,b₁=1,d=2,
∴S_n=a₁+a₂+…+a_n+b₁+b₂+…+b_n
=+n×1+×2
=2ⁿ⁺¹+n²-2.