a(n) = aq^(n-1).|q|正无穷时,s(n) -> a/(1-q) = 所有项之和.
从第(n+1)项以后,所有项之和 = 所有项之和 - 前n项之和 = a/(1-q) - s(n)
a(n) = 2[a/(1-q) - s(n)] = 2[a/(1-q) - a/(1-q) + aq^n/(1-q)] = 2aq^n/(1-q) = [2aq/(1-q)]*q^(n-1)
= aq^(n-1),
2aq/(1-q) = a,
2q/(1-q) = 1,
2q = 1-q,
q = 1/3.
公比q = 1/3.