已知数列{f(n)}的前n项和为sn,且sn=n^2+2n,则:若a1=f(1),a(n+1)=f(an),求{an}的前n项和Tn.
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![已知数列{f(n)}的前n项和为sn,且sn=n^2+2n,则:若a1=f(1),a(n+1)=f(an),求{an}的前n项和Tn.](/uploads/image/z/1770612-60-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Bf%EF%BC%88n%EF%BC%89%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94sn%3Dn%5E2%2B2n%2C%E5%88%99%EF%BC%9A%E8%8B%A5a1%3Df%281%29%2Ca%28n%2B1%EF%BC%89%3Df%28an%29%2C%E6%B1%82%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn.)
已知数列{f(n)}的前n项和为sn,且sn=n^2+2n,则:若a1=f(1),a(n+1)=f(an),求{an}的前n项和Tn.
已知数列{f(n)}的前n项和为sn,且sn=n^2+2n,则:
若a1=f(1),a(n+1)=f(an),求{an}的前n项和Tn.
已知数列{f(n)}的前n项和为sn,且sn=n^2+2n,则:若a1=f(1),a(n+1)=f(an),求{an}的前n项和Tn.
n>=2:f(n)=sn-s(n-1)=2n+1
n=1:s1=1^2+2*1=3=f(1)
故有f(n)=2n+1
f(1)=3=a1
f(an)=2an+1=a(n+1)
等式两边加1,得:a(n+1)+1=2(an+1)
设bn=an+1,数列{bn}是以4为首项,2为公比的等比数列
bn=(b1)×2^(n-1)=2^(n+1)
故an=2^(n+1)-1
Tn=4(1-2^n)/(1-2)-n
=4(2^n-1)-n
f1=3,sn=n/2(f1+fn)=n^2+2n=n/2(4+2n)=n/2(3+2n+1)
fn=2n+1
a1=3
a(n+1)=2an+1
a2=2a1+1
a3=2a2+1=2^2a1+2+1
a4=2a3+1=2^3a1+2^2+2+1
............
an=..........=2^(n-1)a1+2^(n-2)+2^(n-3)+.....+2+1=2^n+2^(n-1)+2^(n-2)+.....+2+1=2^(n+1)-1
Tn=2^(n+2)-n-4
f(n)=Sn-Sn-1=2n+1
a1=3,a(n+1)=2a(n) +1,即[a(n+1) +1]÷[a(n) +1]=2,
因此有a(n) +1= 2的n-1次方 ×(a1+1) =2的n+1次方
a(n) = 2的n+1次方-1
{an}的前n项和Tn=2²-1+2³-1+......+ 2的n+1次方-1
...
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f(n)=Sn-Sn-1=2n+1
a1=3,a(n+1)=2a(n) +1,即[a(n+1) +1]÷[a(n) +1]=2,
因此有a(n) +1= 2的n-1次方 ×(a1+1) =2的n+1次方
a(n) = 2的n+1次方-1
{an}的前n项和Tn=2²-1+2³-1+......+ 2的n+1次方-1
=﹙2²+2³+......2的n+1次方﹚﹣n
=2的n+2次方﹣n-4
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(1):当n=1时,a1=s1=1+2=3
当n≥2时,an=sn-s(n-1)=n²+2n-(n-1)²-2(n-1)=2n+1
当n=1时,a1=3满足an=2n+1
∴通项公式为f(n)=2n+1
(2):a1=f(1)=3,a(n+1)=f(an)=2an+1
∵a(n+1)+1=2(an+1)
∴数列{an+1}是以a1...
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(1):当n=1时,a1=s1=1+2=3
当n≥2时,an=sn-s(n-1)=n²+2n-(n-1)²-2(n-1)=2n+1
当n=1时,a1=3满足an=2n+1
∴通项公式为f(n)=2n+1
(2):a1=f(1)=3,a(n+1)=f(an)=2an+1
∵a(n+1)+1=2(an+1)
∴数列{an+1}是以a1+1=4为首项,以2为公比的等比数列。
Tn=4(1-2^n)/(1-2)-n
=4(2^n-1)-n
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