Question

數列{a_n}的前n項和記為S_n,數列{}是首項為2,公比也為2的等比數列.

(1)求a_n;

(2)若數列{}的前n項和不小于100,問此數列最少有多少項?

Answer 1

解：(1)由題意=2·2^n-1=2ⁿ,

∴S_n=n·2ⁿ.                                                                

當n≥2時,a_n=S_n-S_n-1=n·2ⁿ-(n-1)2^n-1=(n+1)2^n-1,                                    

又當n=1時,a₁=S₁=2,適合上式,

∴a_n=(n+1)2^n-1.                                                               

(2)∵=,

∴數列{}是首項為1,公差為的等差數列,                                

其前n項和為n+n(n-1),

故n+n(n-1)≥100,                                                        

n²+n≥200,得(n+)²≥200+,

滿足它的最小整數是14,即此數列最少有14項.