Question

已知各项都不相等的等差数列{a_n}的前6项和为60,且a₆为a₁和a₂₁的等比中项.

(1)求数列{a_n}的通项公式;

(2)若数列{b_n}满足b_n+1－b_n=a_n(n∈N^*),且b₁=3,求数列的前n项和T_n.

Answer 1

解:(1)设等差数列{a_n}的公差为d(d≠0),则解得∴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=