已知数列an满足a1=12 an 1=an 14n平方-1 ,且a1=1 2
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![已知数列an满足a1=12 an 1=an 14n平方-1 ,且a1=1 2](/uploads/image/f/4267155-3-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D12+an+1%3Dan+14n%E5%B9%B3%E6%96%B9-1+%2C%E4%B8%94a1%3D1+2)
an=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/(a3)+…+1/(an)=2[1/1
a2-a1=2,a3-a2=4,…an+1-an=2n,这n个式子相加,就有an+1=100+n(n+1),即an=n(n-1)+100=n2-n+100,∴ann=n+100n-1≥2n•100n-
你把这个数列看成俩部分a(n1)=2a(n1-1)a(n2)=2n+2an=(an1)+(an2)算算看
2-a(n+1)=12/(an+6)a(n+1)=2an/(an+6)1/a(n+1)=(an+6)/[2an]1/a(n+1)+1/4=3(1/an+1/4)[1/a(n+1)+1/4]/(1/an
an+1=2an+2,an=-1,把an=-1代入bn=2^n/an,得,bn=-2^nb2-b1=-2^*2-(-2)=-6,所以{bn}是等差数列
(1)证b1=a2-a1=1,当n≥2时,bn=an+1−an=an−1+an2−an=−12(an−an−1)=−12bn−1,所以{bn}是以1为首项,−12为公比的等比数列.(2)解由(1)知b
(1)在an+1=3an+1中两边加12:an+12=3(an−1+12),…2分可见数列{an+12}是以3为公比,以a1+12=32为首项的等比数列.…4分故an=32×3n−1−12=3n−12
x=anf(x)=a(n+1)代入函数方程a(n+1)=an^2+2ana(n+1)+1=an^2+2an+1=(an+1)^2满足平方递推数列定义,因此数列{an+1}是平方递推数列.a1+1=10
a[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]=-2^(n-1)-
(Ⅰ)依题意有an+1-1=2an-2且a1-1=2,所以an+1−1an−1=2所以数列{an-1}是等比数列;(Ⅱ)由(Ⅰ)知an-1=(a1-1)2n-1,即an-1=2n,所以an=2n+1而
a(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+3.5^2+...+3.5^n=
(1)∵a1=2,an+1=2an+3.∴an+1+3=2(an+3),a1+3=5∴数列{an+3}是以5为首项,以2为公比的等比数列∴an+3=5•2n−1∴an=5•2n−1−3(2)∵nan=
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)
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)+
a1=2>0假设当n=k(k∈N+)时,ak>0,则a(k+1)=3√ak>0k为任意正整数,因此对于任意正整数n,an恒>0,数列各项均为正.a(n+1)=3√anlog3[a(n+1)]=log3
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
∵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^2a[n]即:a[n+
/>n≥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)(n²-1)an