數(shù)列{an}中,若a1,a2-a1,a3-a2,...,an-an-1,...是首項(xiàng)為1,公比為1/3的等比數(shù)列,則an等于多少?

熱心網(wǎng)友

解：由題意,設(shè)b1=a1=1b2=a2-a1=1/3b3=a3-a2=(1/3)**2……bn=an-a(n-1)=(1/3)**(n-1)q=1/3則，b1+b2+b3+…+bn=a1+(a2-a1)+(a3-a2)+…+(an-a(n-1))=an且，b1+b2+b3+…+bn=1+1/3+(1/3)**2+(1/3)**(n-1)=1/(1-q)=1/(1-1/3)=1/(2/3)=3/2所以，an=3/2

熱心網(wǎng)友

解:(a2 - a1)/a1 = 1/3，a2 = (4/3)^1(a3 - a2)/a2 = 1/3，a3 = (4/3)^2(a4 - a3)/a3 = 1/3，a4 = (4/3)^3……(ak - a(k - 1))/a(k - 1) = 1/3，ak = (4/3)^(k - 1)……(an - a(n - 1))/a(n - 1) = 1/3，an = (4/3)^(n - 1)

熱心網(wǎng)友

解：由提議：a1=1a2-a1=(1/3)^1a3-a2=(1/3)^2...a(n-1)-a(n-2)=(1/3)^(n-2)an-a(n-1)=(1/3)^(n-1)以上各式相加：an=[(1/3)^n-1]/(1/3-1)=3/2-(1/3)^(n-1)/2

熱心網(wǎng)友

解：由已知可以得到an-a(n-1)=(1/3)^(n-1)a(n-1)-a(n-2)=(1/3)^(n-2)a(n-2)-a(n-3)=(1/3)^(n-4).....................a2-a1=(1/3)^(2-1)a1=1把這n個(gè)等式的兩邊分別相加，得到an=(1/3)^(n-1)+(1/3)^(n-2)+(1/3^(n-3)+......+1/3+1=[[(1/3)^n-1]/(1/3-1)=2-2(1/3)^n, (n=2)an=1 (n=1)