Question

已知數(shù)列{a_n}滿足a_n=2a^n-1+2ⁿ+2(n≥2),a₁=2.

(1)求a₂,a₃,a₄;

(2)是否存在一個(gè)實(shí)數(shù)λ,使得數(shù)列{}成等差數(shù)列,若存在,求出λ的值;若不存在,請(qǐng)說(shuō)明理由;

(3)設(shè)S_n為數(shù)列{a_n}的前n項(xiàng)和,證明S_n＞n³+n².

Answer 1

解:(1)a₂=4+4+2=10,a₃=20+8+2=30,a₄=60+16+2=78.

(2)假設(shè)存在一個(gè)實(shí)數(shù)λ,使得數(shù)列{}成等差數(shù)列,則==1+恒為常數(shù),

∴2-λ=0,即λ=2.而+=1,

∴λ=2時(shí)數(shù)列{}為等差數(shù)列.

(3)解法一:=+(n-1)=n+1,

∴a_n=(n+1)2ⁿ-2.

S_n=2·2+3·2²+4·2³+…+(n+1)2ⁿ-2n,

2S_n=2·2²+3·2³+4·2⁴+…+(n+1)2ⁿ⁺¹-4n.

兩式相減得

-S_n=2·2+2²+2³+…+2ⁿ-(n+1)2ⁿ⁺¹+2n=-n·2ⁿ⁺¹+2n.

∴S_n=n·2ⁿ⁺¹-2n=2n(2ⁿ-1)=2n［(1+1)ⁿ-1］=2n(1+n++…-1)≥n³+n².

解法二:用數(shù)學(xué)歸納法也可.