Question

數(shù)列{a_n}的前n項(xiàng)和記為S_n,數(shù)列{}是首項(xiàng)為2,公比也為2的等比數(shù)列.

(1)求a_n;

(2)若數(shù)列{}的前n項(xiàng)和不小于100,問此數(shù)列最少有多少項(xiàng)?

Answer 1

解：(1)由題意=2·2^n-1=2ⁿ,

∴S_n=n·2ⁿ.                                                                

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

又當(dāng)n=1時(shí),a₁=S₁=2,適合上式,

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

(2)∵=,

∴數(shù)列{}是首項(xiàng)為1,公差為的等差數(shù)列,                                

其前n項(xiàng)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(shù)是14,即此數(shù)列最少有14項(xiàng).