Question

已知数列{an}的前n项和Sn=pn+q（p≠0且p≠1）,求证：数列{an}为等比数列的充要条件为q=-1．

Answer 1

先证明必要性

当q = -1时,S = p^n - 1 ,

a = S - S = p^n -1 -[p^(n-1) - 1] = p^n - p^(n-1) ,

a/a = [p^(n+1) - p^n]/[p^n - p^(n-1)] = (p -1)/(1-1/p) = p ≠ 0,所以数列{a}为等比数列

再证其充分性

S = p^n + q ,a = S - S = p^n+q-[p^(n-1)+q] = p^n - p^(n-1)

由上知 a/a = p ≠ 0 ,所以 a/a = p

a = p + q ,a = S - a = p^2 + q - (p + q) = p^2 - p

a/a = (p^2 - p)/(p + q) = p ,由p≠0且p≠1,所以q = -1.