已知数列an的各项满足:a1=1-3k,an=4^n-1-3a(n-1
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1,已知数列a‹n›各项为正数,a₁≠2,且前n项之和满足6S‹n›=a‹n›²+3a‹n
∵Sn-Sn-1=√Sn+√Sn-1∴(√Sn)²-(√Sn-1)²=√Sn+√Sn-1(√Sn-√Sn-1)(√Sn+√Sn-1)=√Sn+√Sn-1∴√Sn-√Sn-1=1(n
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
那么我把Aˇ〔3/2〕n+1理解成A[n+1]的3/2次方了递推式可以化成A[n]/A[n+1]^2=(A[n+1]/A[n+2]^2)^(-1/2)两边取对数得到log(A[n]/A[n+1]^2)
由题意,Sn=n^2,则a1=1,S(n-1)=(n-1)^2=n^2-2n+1,n>=2an=Sn-S(n-1)=n^2-n^2+2n-1=2n-1,n>=2由于当n=1时,2n-1=1=a1所以,
an>0n=1时S1=a1=(a1²+a1)/2∴a1=1n>=2时S(n-1)=(a(n-1)²-a(n-1))/2an=Sn-S(n-1)∴(an+a(n-1))(an-a(n
a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(
(I)由a1=S1=1/6(a1+1)(a1+2),解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,又由a(n+1)=S(n+1)-Sn=1/6(a(n+1)+1)(a(n+1)+2)-1
[2a(n+1)-an]/[2an-a(n+1)]=ana(n+1)2an²a(n+1)-ana(n+1)²=2a(n+1)-an2an²a(n+1)-2a(n+1)=a
化简等式[a(n+1)-a(n)]²+1=2[a(n+1)+a(n)][a(n+1)-a(n)+1]²=2[a(n+1)+a(n)]+2[a(n+1)-a(n)][a(n+1)-a
设前n项和为Sn,Sn=n的平方,那么前(n-1)项S(n-1)的和为(n-1)的平方.Sn-S(n-1)=an{an}的通项就是n的平方减(n-1)的平方结果是2n-1哎呀我的妈呀不会打n的平方累死
1.an=Sn-S(n-1)=2n^2-3n-2(n-1)^2+3(n-1)=4n-5a1=-1b1=-a1=1a2=3b3(a2-a1)=b3(3+1)=1b3=1/4=b1q^2=q^2q=1/2
1.a_(1)=1,a_(n+1)=2a_(n)+2^(n)----------------1b_(n)=a_(n)/2^(n)将式子1左右两边同时除以2^(n+1),则:b_(n+1)=b_(n)+
d(n)=2^n+n,p(1)=d(1)=2^1+1=3,p(n+1)=d(n+1)+d(n)=2^(n+1)+(n+1)+2^n+n=3*2^n+2n+1,L(2n-1)=d(2n-1)=2^(2n
an=a(n-1)+a(n-2)+……+a2+a1所以a2=a1=8而且当n>1时,an=S(n-1)又有an=Sn-S(n-1)=a(n+1)-an2an=a(n+1)所以n>1的部分是等比数列,公
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)
an=4^n-1-3a(n-1)an-4^n/7=-3[a(n-1)-(4^n-1)/7][an-4^n/7]/[a(n-1)-(4^n-1)/7]=-3成等比数列所以an=a1*(-3)^(n-1)
1.A(n+1)^2*An+A(n+1)*An^2+A(n+1)^2-An^2=0两边同除以A(n+1)²An²1/An+1/A(n+1)+1/An²-1/A(n+1)&
a1=10an=9*10的n-1次方
a(3)=a(1+2)=1/[1+a(1)]=a(1),1=a(1)+[a(1)]^2,0=[a(1)]^2+a(1)-1,Delta=1+4=5.a(1)=[-1+5^(1/2)]/2,或a(1)=