设｛a下n｝是公比大于1的等比数列,s下n为其前n项和,已知s下3=7,且a下1+3,3a下2,a下3+4成等差数列求数列｛a下n｝的通项公式设b下n=lna下3n加1,n∈N上＊,求数列｛b下n｝的前n项和T下n

Question

设｛a 下n｝是公比大于1的等比数列,s下n为其前n项和,已知s下3=7,且a下1+3,3a下2,
a下3+4成等差数列 求数列｛a下n｝的通项公式 设b下n=ln a下3n加1,n∈N上＊,求数列｛b下n｝的前n项和T下n

Answer 1

a(n) = aq^(n-1), q>1.
s(n) = a[q^n - 1]/(q-1).
7 = s(3) = a(1 + q + q^2),
2[3a(2)] = [a(1)+3] + [a(3)+4] = a+3+aq^2+4 = 6aq,
7 + a + aq^2 = 6aq,
7 + a + aq + aq^2 = 7aq = 7 + a(1+q+q^2) = 7 + 7 = 14.
aq = 2. a = 2/q.
7 = a(1+q+q^2) = 2(1+q+q^2)/q,
0 = 2q^2 + 2q + 2 - 7q = 2q^2 - 5q + 2 = (2q-1)(q-2),
q = 1/2 < 1,或者, q = 2.
因q>1.因此,只有q=2.
a=2/q = 1.
a(n) = 2^(n-1).
b(n) = ln[a(3n+1)] = ln[2^(3n)] = ln(8^n) = nln(8).
t(n) = b(1)+b(2)+...+b(n) = ln(8)[1+2+...+n] = n(n+1)/2*ln(8) = ln(8)n(n+1)/2