設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且對任意正整數(shù)n,an+Sn=4096.
(1)求數(shù)列{an}的通項(xiàng)公式
(2)設(shè)數(shù)列{log2an}的前n項(xiàng)和為Tn,對數(shù)列{Tn},從第幾項(xiàng)起Tn<-509?
【答案】
分析:(1)由題設(shè)知a
1+S
1=4096,a
1=2048.a(chǎn)
n=S
n-S
n-1=(4096-a
n)-(4096-a
n-1)=a
n-1-a
n,由此可知
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=
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a
n=2048(
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)
n-1.
(2)由題設(shè)條件可知T
n=
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(-n
2+23n).再由T
n<-509,知n>
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,由此可知從第46項(xiàng)起T
n<-509.
解答:解:(1)∵a
n+S
n=4096,
∴a
1+S
1=4096,a
1=2048.
當(dāng)n≥2時,a
n=S
n-S
n-1=(4096-a
n)-(4096-a
n-1)=a
n-1-a
n∴
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=
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a
n=2048(
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)
n-1.
(2)∵log
2a
n=log
2[2048(
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)
n-1]=12-n,
∴T
n=
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(-n
2+23n).
由T
n<-509,解得n>
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,而n是正整數(shù),
于是,n≥46.
∴從第46項(xiàng)起T
n<-509.
點(diǎn)評:本題考查數(shù)列的綜合應(yīng)用,解題時要認(rèn)真審題,仔細(xì)解答.