Question

设{a_n}是公比不为1的等比数列,其前n项和为S_n,且a₅,a₃,a₄成等差数列.

(1)求数列{a_n}的公比.

(2)证明:对任意k∈N^*,S_k+2,S_k,S_k+1成等差数列.

Answer 1

【解析】(1)设数列{a_n}的公比为q(q≠0,q≠1),

由a₅,a₃,a₄成等差数列,得2a₃=a₅+a₄,即2a₁q²=a₁q⁴+a₁q³,

由a₁≠0,q≠0得q²+q-2=0,解得q₁=-2,q₂=1(舍去),

所以q=-2.

(2)对任意k∈N^*,

S_k+2+S_k+1-2S_k=(S_k+2-S_k)+(S_k+1-S_k)

=a_k+1+a_k+2+a_k+1=2a_k+1+a_k+1·(-2)=0,

所以对任意k∈N^*,S_k+2,S_k,S_k+1成等差数列.