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数学
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数列{an}中an+1=c*an(c为非零常数)前n项和Sn=3^n+k,k=
A)-1
(B)1
(C)0
(D)2
人气:270 ℃ 时间:2020-04-07 11:37:55
解答
Sn+1=3^(n+1)+k
Sn=3^n+k
a(n+1)=Sn+1-Sn=2*3^n
所以 a1=2
因为an+1/an=c
所以an是等比数列
由an=2*3^(n-1)可得c=3
Sn=2*(1-3^n)/(1-3)=3^n-1
所以 k=-1
选A
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