A1=1,An 1=3An² 2 求An
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a(n+1)=3an/(an+3)a2=(3*1/2)/(1/2+3)=(3/2)/(7/2)=3/7a3=(3*3/7)/(3/7+3)=(9/7)/(24/7)=9/24=3/8a4=(3*3/8
由于a1=-2,an+1=1−an1+an∴a2=1+a11−a1=−13,a3=1+a21−a2=12,a4=1+a31−a3=3,a5=1+a41−a4=−2=a1∴数列{an}以4为周期的数列∴
由an+2=3an+1-2an可得an+2-an+1=2(an+1-an)因为a2-a1=2,所以an+1-an不会等于0,则an+1-an是以2为公比的等比数列由上可得an+1-an=2^nan-a
方法1:an+1=an/(2an+3)两边取倒数:1/a(n+1)=2+3/an设bn=1/anb(n+1)=3bn+2b(n+1)+1=3(bn+1)[b(n+1)+1]/(bn+1)=3∴{bn+
A(n+1)-1/3=2(A(n)-1/3)B(n)=A(n)-1/3B(n)=B(1)*2~(n-1)=(5/3)*2~(n-1)A(n)=(5/3)*2~(n-1)+1/3
等式两边倒数,得到1/an+1=1+3/an,再变形,得到:(1/an+1)+1/2=3(1/an+1/2)所以{bn}={1/an+1/2}是一个等比数列,第一项b1=1/a1+1/2=1所以bn=
an-1=2an+3an-1+3=2(an+3)(an+3)/(an-1+3)=1/2所以an+3为等比数列an+3=(a1+3)*q^(n-1)=4*(1/2)^(n-1)
跟你说一般方法吧:通项公式是:A(n+1)+k=b(An+k),即A(n+1)-bAn+k(1-b)=0,对照原式解出k和b,其余步骤跟上题相同.
a(n+1)-an=3*2^(2n-1)所以:an-a(n-1)=3*2^(2n-3)...a3-a2=3*2^3a2-a1=3*2^1上述各项相加:an-a1=3[2^1+2^3+2^5+2^7+.
a1=1an=an-1+3n-2an-1=an-2+3(n-1)-2...a2=a1+3*2-2左右分别相加an=a1+3*(n+n-1+...+2)-2*(n-1)an=1+3*(n+2)*(n-1
(1)an=2a+3,∴an+3=2[a+3],∴数列{an+3}是等比数列.(2)an+3=(a1+3)*2^(n-1),an=(a1+3)*2^(n-1)-3=(6)*2^(n-1)-3.再问:2
a(n+1)=2an/(3an+4)化成1/a(n+1)=(3an+4)/2an=3/2+2/an
(1)证明:若an+1=an,即2an1+an=an,解得an=0或1.从而an=an-1=…a2=a1=0或1,与题设a1>0,a1≠1相矛盾,故an+1≠an成立.(2)由a1=12,得到a2=2
待定系数法因为a(n+1)=2an-n^2+3n设a(n+1)+p(n+1)^2+q(n+1)=2(an+pn^2+qn)展开整理得a(n+1)=2an+pn^2+(q-2p)-(p+q)与原式一一对
an=3n-1由an+1=an+3得知公差d=3所以an=a1+(n-1)d=3n-1
依次第二列加上第一列,第三列加上第二列...原式=-a100...00-a20...0.000...-an0123...nn+1所以原式=(n+1)*(-1)^n*a1*a2*...*an
a(n+1)=2an+3a(n+1)+k=2an+3+k=2(an+3/2+k/2)则令k=3/2+k/2k=3则两边同时加3a(n+1)+3=2(an+3)[a(n+1)+3]/(an+3)=2所以
是等比数列吧?3a(n+1)-an=03a(n+1)=ana(n+1)/an=1/3,等比1/3a1=2an=2/3^(n-1)=6/3^n
2an+1=an+1得2(an+1)=an+1+1a1=3a1+1=4a2+1=8则an+1=2^(n+1)an=2^(n+a)-1
∵1=2,an+1=1+an1−an(n∈N*),∴a2=1+a11−a1=1+21−2=-3,a3=1+a21−a2=1−31+3=−12a4=1+a31−a3=1−121+12=13a5=1+a4