Question

数列{a_n}的前n项和记为S_n,数列{}是首项为2,公比也为2的等比数列.

(1)求a_n;

(2)若数列{}的前n项和不小于100,问此数列最少有多少项?

Answer 1

解：(1)由题意=2·2^n-1=2ⁿ,

∴S_n=n·2ⁿ.                                                                

当n≥2时,a_n=S_n-S_n-1=n·2ⁿ-(n-1)2^n-1=(n+1)2^n-1,                                    

又当n=1时,a₁=S₁=2,适合上式,

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

(2)∵=,

∴数列{}是首项为1,公差为的等差数列,                                

其前n项和为n+n(n-1),

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

n²+n≥200,得(n+)²≥200+,

满足它的最小整数是14,即此数列最少有14项.