已知数列an满足a1=1

已知数列{an}满足a1=1an+1=an/(3an+1)则球an求通项公式倒数变形经常可以尝试一下:

已知数列{an}满足an+1=an+n,a1等于1,则an=?A2=A1+1A3=A2+2A4=A3+3.An=A(n-1)+(N-1)左式上下相加=右式上下相加An=A1+[1+2+3+...+(N-1)]An=1+[N(N-1)]/2你

已知数列{an}满足an+1=2an+3.5^n,a1=6.求ana(n+1)-2an=3.5^n,则a2-2a1=3.5^1a3-2a2=3.5^2.a(n+1)-2an=3.5^n以上式子相加,得a(n+1)-a1-Sn=3.5+3.5

已知数列{an}满足a1=4,an=4-4/an-1已知数列{an}满足a1=4,an=4-4/a（n-1）(n≥2),令bn=1/an-2.1、求证：数列{bn}是等差数列2、求数列{an}通项an=4-4/a(n-1)an-2=2-4/

已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)1/a(n+

已知数列An满足A1=1,且4An+1-AnAn+1+2An=9已知数列An满足A1=1,且4An+1-AnAn+1+2An=9猜想An的通向公式并用数学归纳法证明∵数列{a[n]}满足4a[n+1]-a[n]a[n+1]+2a[n]=9∴

已知数列{an}满足a(n+1)=an+n,a1=1,则an=a(n+1)=an+n即a(n+1)-an=n所以n>=2时有an-a(n-1)=n-1.a2-a1=1以上n-1个式子相加得到an-a1=1+2+3+...+（n-1）=n*(

已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/anan=1＋2＋3＋…＋n=[n(n＋1)]/2则：1/(an)=2/[n(n＋1)]=2[(1/n)－1/(n＋1)],所以：M=1/(a1)＋1/(a2)＋1

已知数列满足a1=1,an+1+2an=2求ana(n+1)+2an=1a(n+1)-1/3=-2an+2/3=-2(an-1/3)所以,{an-1/3}是以a1-1/3=2/3为首项,公比是-2的等比数列.即an-1/3=2/3*(-2)

已知数列{an}满足a1=1,an+1·an=2^n则s2012我来帮你解答吧!O(∩_∩)O~　　由an+1·an=2^n,可得an+2·an+1=2^（n+1）,两者相除,可得an+2/an=2,观察此式,可以发现,数列{an}的奇数项

已知数列{an}满足An+1=2^nAn,且A1=1,则通项an解An+1/An=2^n所以A2/A1=2所以数列是以1为首相2为公比的等比数列所以通向公式an=2^(n-1)An+1/An=2^nA1=2^0A2=2^1A3=2^(1+2

已知数列an满足条件a1=-2an+1=2an+1则a5a[n+1]=2a[n]+1a[n+1]+1=2(a[n]+1)则{a[n]+1}是公比为2的等比数列a[1]+1=-2+1=-1所以a[n]+1=(-1)*2^(n-1)a[n]=-

已知数列{an}满足a(n+1)=an+lg2,a1=1,求ana(n+1)-an=lg2an-a(n-1)=lg2a(n-1)-a(n-2)=lg2...a3-a2=lg2a2-a1=lg2累加得an-a1=(n-1)lg2所以an=(n

已知数列｛an｝满足an=an+1-lg2,且a1=1,则通项公式为?a(n+1)=a(n)+lg2a(n)=a(n-1)+lg2a(n-1)=a(n-2)+lg2...a(2)=a(1)+lg2a(n+1)-a(1)+(a(n)+a(n-

已知数列{an},满足a1=1/2,Sn=n²×an,求ann≥2时,Sn=n²×anS(n-1)=(n-1)²×a(n-1)an=Sn-S(n-1)=n²×an-(n-1)²×a(n-1)

已知数列{an}满足a1=1/2,sn=n^2an,求通项an∵s[n]=n^2a[n]∴s[n+1]=(n+1)^2a[n+1]将上述两式相减,得：a[n+1]=(n+1)^2a[n+1]-n^2a[n](n^2+2n)a[n+1]=n^

已知数列{an}满足3an+1+an=4,a1=9,求通项公式.3a(n+1)=-an+43a(n+1)-3=-an+13[a(n+1)-1]=-(an-1)[a(n+1)-1]/(an-1)=-1/3所以an-1是等比数列,q=-1/3a

已知数列an满足a1=1/2sn=n平方×an求an如图an=Sn-S(n-1)=n^2*an-(n-1)^2*a(n-1)(n^2-1)an=(n-1)^2*a(n-1)(n+1)(n-1)an=(n-1)^2*a(n-1)an=(n-1

已知数列An满足A1=1/2Sn=N²An求An易得a2=1/6、a3=1/12、a4=1/20；猜想an=1/[n(n+1)]数学归纳法证明:①当n=1时,a1=1/2成立；②假设n=k(k≥2)时,ak=1/[k(k+1)]成

已知数列an满足an+1=nan,a1=2,求an通项a(n+1)=nana(n+1)/an=n那么有:an/a(n-1)=n-1...a2/a1=1以上各项相乘得:an/a1=1*2*..(n-1)=(n-1)!所以,an=a1*(n-1