1、设等比数列{an}的公比为q,前n项和为Sn >0 (n=1、2、3…)(1) 求q的取值范围
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1、设等比数列{an}的公比为q,前n项和为Sn >0 (n=1、2、3…)(1) 求q的取值范围
(2)设bn=a下标(n+2)-1.5a下标(n+1) ,记{bn}的前n项和为Tn,试比较Sn和Tn的大小.(不懂做,Sn=Tn)
(2)设bn=a下标(n+2)-1.5a下标(n+1) ,记{bn}的前n项和为Tn,试比较Sn和Tn的大小.(不懂做,Sn=Tn)
a(n) = aq^(n-1),
a = a(1) = S(1) > 0,
q = 1时,S(n) = na > 0.满足要求.
q不等于1时,
S(n) = a[q^n-1]/(q-1).
q>1时,q^n-1>0,q-1>0,S(n) = a[q^n-1]/(q-1) >0.满足要求.
-1-1.
b(n) = a(n+2) - 1.5a(n+1) = aq^(n+1) - 1.5aq^n = aq^n[q-1.5].
q = 1时,b(n) = a(-0.5),T(n) = -na/2,S(n) = na > -na/2 = T(n).
q > -1且q不等于1时,T(n) = aq(q-1.5)[q^n-1]/(q-1),S(n) = a[q^n-1]/(q-1).
T(n) - S(n) = a[q^n-1]/(q-1)[q(q-1.5) - 1] = a[q^n-1][2q^2 - 3q - 2]/[2(q-1)] = a[q^n-1][2q+1][q-2]/[2(q-1)]
-1 < q < -1/2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] > 0,
T(n) > S(n).
q = -1/2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] = 0,
T(n) = S(n).
-1/2 < q < 1时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] < 0,
T(n) < S(n).
1 < q < 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] < 0,
T(n) < S(n).
q = 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] = 0,
T(n) = S(n).
q > 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] > 0,
T(n) > S(n).
综合,有
-1 < q < -1/2时,T(n) > S(n).
q = -1/2时,T(n) = S(n).
-1/2 < q < 2时,T(n) < S(n).
q = 2时,T(n) = S(n).
q > 2时,T(n) > S(n).
a = a(1) = S(1) > 0,
q = 1时,S(n) = na > 0.满足要求.
q不等于1时,
S(n) = a[q^n-1]/(q-1).
q>1时,q^n-1>0,q-1>0,S(n) = a[q^n-1]/(q-1) >0.满足要求.
-1-1.
b(n) = a(n+2) - 1.5a(n+1) = aq^(n+1) - 1.5aq^n = aq^n[q-1.5].
q = 1时,b(n) = a(-0.5),T(n) = -na/2,S(n) = na > -na/2 = T(n).
q > -1且q不等于1时,T(n) = aq(q-1.5)[q^n-1]/(q-1),S(n) = a[q^n-1]/(q-1).
T(n) - S(n) = a[q^n-1]/(q-1)[q(q-1.5) - 1] = a[q^n-1][2q^2 - 3q - 2]/[2(q-1)] = a[q^n-1][2q+1][q-2]/[2(q-1)]
-1 < q < -1/2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] > 0,
T(n) > S(n).
q = -1/2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] = 0,
T(n) = S(n).
-1/2 < q < 1时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] < 0,
T(n) < S(n).
1 < q < 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] < 0,
T(n) < S(n).
q = 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] = 0,
T(n) = S(n).
q > 2时,T(n) - S(n) = a[q^n-1][2q+1][q-2]/[2(q-1)] > 0,
T(n) > S(n).
综合,有
-1 < q < -1/2时,T(n) > S(n).
q = -1/2时,T(n) = S(n).
-1/2 < q < 2时,T(n) < S(n).
q = 2时,T(n) = S(n).
q > 2时,T(n) > S(n).
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