已知各项为正数an满足a1等于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
a1^2+a2^2+a3^2+……an-1^2=(4(n-1/)^3-(n-1))/3a1^2+a2^2+a3^2+……an^2=(4n^3-n)/3两式相减可得an^2=(2n-1)^2所以an=2
∵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
a1=1,a2=q,a3=q^2,则a1+a2+a3=1+q+q^2=7,即q^2+q-6=0,解得q=2或q=-3(舍去),所以q=2,所以an=a1×q^(n-1)=2^(n-1)
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)
∵a1,1/2a3,2a2成等差数列∴2×1/2a3=a1+2a22即a3=a1+2a2∵{an}是等比数列,∴a1q²=a1+2a1q∴q²=1+2q,即q²-2q-1
解因为数列是等比数列,且公比为q则a2=a1qa3=a1q²又因为a1,1/2a3,2a2成等差数列所以有2*(1/2)a3=a1+2a2即a1q²=a1+2a1q即q²
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
∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n
a1(q+q^3)=4a1(1+q+q^2)=14两式相除:(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=(a1+a2+a3)*q^(n-1)>1/9关键是求q说实在的,我
可用递推法:2Sn=An+An*An递推2Sn-1=An-1+An-1*An-1两市相减,得:An+An-1=An*An-An-1*An-1因为An为正数,所以An-An-1=1之后求An,然后用求和
设等比数列的公比为q,由2a1,12a3,a2成等差数列,得a3=2a1+a2,即a1q2=2a1+a1q=a1(2+q),因为a1≠0,所以q2=2+q,解得q=-1或q=2.因为等比数列{an}各
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)
设公比为q,首项为a1,则由a1,二分之一a3,a2成等差数列可得/a3=a1+a2即a1*q^2=a1+a1*qq^2=1+q可求得q=(1+√5)/2(a3+a2)/(a4+a5)=(a1*q^2
^代表什么的几次方a1=1,设等比为q且q〉0,则a1+a1*q+a1*q^2=14即a1*(1+q+q^2)=14将a1代入得q^2+q-6=0解得q=-3(舍去)q=2通过验证an=2*2^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)&
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)=