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的等比數(shù)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
等比數(shù)列前n項(xiàng)和Sn=k+2*(1/3)∧n,則常數(shù)k的值為?
等比數(shù)列前n項(xiàng)和Sn=k+2*(1/3)∧n,則常數(shù)k的值為?
數(shù)學(xué)人氣：739 ℃時(shí)間：2020-04-08 19:51:55
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