Question

已知各項(xiàng)都不相等的等差數(shù)列{a_n}的前6項(xiàng)和為60,且a₆為a₁和a₂₁的等比中項(xiàng).

(1)求數(shù)列{a_n}的通項(xiàng)公式;

(2)若數(shù)列{b_n}滿足b_n+1－b_n=a_n(n∈N^*),且b₁=3,求數(shù)列的前n項(xiàng)和T_n.

Answer 1

解:(1)設(shè)等差數(shù)列{a_n}的公差為d(d≠0),則解得∴a_n=2n+3.

(2)由b_n+1-b_n=a_n,∴b_n-b_n-1=a_n-1(n≥2,n∈N^*),

b_n=(b_n-b_n-1)+(b_n-1-b_n-2)+…+(b₂-b₁)+b₁=a_n-1+a_n-2+…+a₁+b₁=(n-1)·+3=n(n+2).

∴b_n=n(n+2)(n∈N^*).

∴,

T_n=