等比数列前n项和Sn=k+2*(1/3)∧n,则常数k的值为?
问题描述：
等比数列前n项和Sn=k+2*(1/3)∧n,则常数k的值为?
答
Sn=k+2*(1/3)∧n
S(n-1)=k+2*(1/3)∧(n-1)
An=Sn-S(n-1)=2*(1/3)∧(n-1)(1/3-1)
=2*(1/3)∧(n-1)*2/3
=(1/3)∧(n-1)*1/3
=(1/3)∧n
所以是以公比q=1/3的等比数列
Sn=A1(q∧n-1)/(q-1)
=1/3((1/3)∧n-1)/(1/3-1)
=1/3(1-(1/3)∧n)/2/3
=1/2-1/2*(1/3)∧n=k+2*(1/3)∧n
1-(1/3)∧n=2k+4*(1/3)∧n
2k=1-5*(1/3)∧n
k=(1-5*(1/3)∧n)/2