设数列{an}是各项均为正数的等比数列,Sn为其前n项和,m、n、p均为正整数,且满足m+n=2p,求证:1S2m+1S2n≥2S2p.-高二数学

题目简介

设数列{an}是各项均为正数的等比数列,Sn为其前n项和,m、n、p均为正整数,且满足m+n=2p,求证:1S2m+1S2n≥2S2p.-高二数学

题目详情

设数列{an}是各项均为正数的等比数列,Sn为其前n项和,m、n、p均为正整数,且满足m+n=2p,求证:
1
S2m
+
1
S2n
2
S2p
题型:解答题难度:中档来源:不详

答案

当各项均为正数的等比数列{an}的公比q=1时,class="stub"1
S2m
+class="stub"1
S2n
=class="stub"1
(ma1)2
+class="stub"1
(na1)2
=class="stub"1
a12
class="stub"1
m2
+class="stub"1
n2
)≥class="stub"1
a12
×class="stub"2
mn

∵m、n、p均为正整数,且满足m+n=2p,
∴2p≥2
mn

class="stub"2
p2
class="stub"2
mn

class="stub"1
a12
class="stub"2
p2
class="stub"1
a12
class="stub"2
mn
,又class="stub"2
S2p
=class="stub"1
a12
class="stub"2
p2

class="stub"1
S2m
+class="stub"1
S2n
class="stub"2
S2p

当q≠1时,class="stub"1
S2m
=
(1-q)2
a12(1-qm)2
class="stub"1
S2n
=
(1-q)2
a12(1-qn)2
class="stub"1
S2p
=
(1-q)2
a12(1-qp)2

要证class="stub"1
S2m
+class="stub"1
S2n
class="stub"2
S2p
,只需证class="stub"1
(1-qm)2
+class="stub"1
(1-qn)2
class="stub"2
(1-qp)2

class="stub"1
(1-qm)2
+class="stub"1
(1-qn)2
class="stub"2
(1-qm)(1-qn)

∴只需证(1-qm)•(1-qn)≤(1-qp)2,
即证-qm-qn+qm+n≤-2qp+q2p,∵m+n=2p,
∴只需证qm+qn≥2qp.
∵qm+qn≥2
qm•qn
=2
qm+n
=2qclass="stub"m+n
2
=2qp成立,
∴q≠1时,原结论成立.
综上所述,class="stub"1
S2m
+class="stub"1
S2n
class="stub"2
S2p

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