设首项为a,公差为d的等差数列前n项的和为An,又首项为a,公比为r的等比数列前n项和为Gn,其中a≠0,|r|<1.令Sn=G1+G2+…+Gn,若有limn→∞(Ann-Sn)=a,求r的值.-数

题目简介

设首项为a,公差为d的等差数列前n项的和为An,又首项为a,公比为r的等比数列前n项和为Gn,其中a≠0,|r|<1.令Sn=G1+G2+…+Gn,若有limn→∞(Ann-Sn)=a,求r的值.-数

题目详情

设首项为a,公差为d的等差数列前n项的和为An,又首项为a,公比为r的等比数列前n项和为Gn,其中a≠0,|r|<1.令Sn=G1+G2+…+Gn,若有
lim
n→∞
(
An
n
-Sn)
=a,求r的值.
题型:解答题难度:中档来源:不详

答案

由题意知Gn=
a(1-rn)
1-r

∴Sn=class="stub"1
-1+r
•[a(r +r2+r3…+rn)-(a+a+a…+a)]

=class="stub"1
-1+r
(
ar(1-rn)
1-r
-na)

=class="stub"a
(-1+r)2
[rn-r-n(-1+r)]
An=na+
n(n-1)
2
•d

An
n
-Sn
=class="stub"1
n
[na+
n(n-1)
2
•d
]-class="stub"a
(-1+r)2
[rn-r-n(-1+r)]=a+class="stub"n-1
2
•d
-class="stub"a
(-1+r)2
×(rn-r)-class="stub"an
1-r

lim
n→∞
(
An
n
-Sn)
=a,a≠0,|r|<1
所以:class="stub"d
2
+class="stub"a
r-1
=0且class="stub"a
(1-r)2
×r+a-class="stub"d
2
=a,即class="stub"a
(1-r)2
×r-class="stub"d
2
=0
class="stub"a
(1-r)2
×r+class="stub"a
r-1
=0,整理得2r-1=0,解得r=class="stub"1
2

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