等比数列一道题(An)的前n项和Sn满足Sn=(a/(a-1))(An-1) (a为常数a不等0和1)求正 (An)是等比数列
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等比数列一道题(An)的前n项和Sn满足Sn=(a/(a-1))(An-1) (a为常数a不等0和1)求正 (An)是等比数列
等比数列一道题
(An)的前n项和Sn满足Sn=(a/(a-1))(An-1) (a为常数a不等0和1)
求正 (An)是等比数列
等比数列一道题(An)的前n项和Sn满足Sn=(a/(a-1))(An-1) (a为常数a不等0和1)求正 (An)是等比数列
Sn=(a/(a-1))(An-1)
S(n-1)=(a/(a-1))(A(n-1)-1)
相减得:
An=(a/(a-1)/[An-A(n-1)]
An-a/(a-1)An=-a/(a-1)A(n-1)
An=a*A(n-1)
故{An}是以a为公比的等比数列
LS的过程有误。
证明:
由于:Sn=[a/(a-1)]*[(An)-1]
则有:S(n-1)=[a/(a-1)]*[A(n-1)-1]
则:n>=2时
An=Sn-S(n-1)
=[a/(a-1)]*[An-1-(A(n-1)-1)]
=[a/(a-1)]*[An-A(n-1)]
则:[a/(a-1)]A(n-...
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LS的过程有误。
证明:
由于:Sn=[a/(a-1)]*[(An)-1]
则有:S(n-1)=[a/(a-1)]*[A(n-1)-1]
则:n>=2时
An=Sn-S(n-1)
=[a/(a-1)]*[An-1-(A(n-1)-1)]
=[a/(a-1)]*[An-A(n-1)]
则:[a/(a-1)]A(n-1)=[1/(a-1)]An
a*A(n-1)=An
An/A(n-1)=a
又由于:
A2/A1=a
则:{An}是公比为a等比数列
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