数列an满足a1=2,an 1-an=3*2 2n-1
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我给你求出来吧an+1=an/(2an+3)两边取倒数1/an+1=(2an+3)/an=2+3/an设1/an=bn则bn+1=3bn+2所以1+bn+1=3(1+bn)所以{1+bn}等比数列首项
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
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,
a(n+1)=3an+1a(n+1)+1/2=3an+3/2=3(an+1/2)[a(n+1)+1/2]/(an+1/2)=3,为定值.a1+1/2=1/2+1/2=1数列{an+1/2}是以1为首项
(1)根据题意,有An=(An-An-1)+(An-1-An-2)+…+(A2-A1)+A1=3-2^(2n-3)+3-2^(2n-5)+…+(3-2^3)+2再用分组求和法:=3n-【2^(2n-3
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
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
A1=1/2成立,设An=1/[n(n+1)]成立,因为A1+A2+…+An=n^2An所以A1+A2+…+An+A(n+1)=(n+1)^2A(n+1),所以A(n+1)=(n+1)^2A(n+1)
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)-
(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
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=
a(n+1)=4an+9(n+1)表示下标a(n+1)+3=4(an+3)[a(n+1)+3]/(an+3)=4所以数列{an+3}是以a1+3=5为首相q=4为公比的等比数列an+3=5*(4)^(
你应该是抄错题了吧--A(n+1)=2An+2^n等式两边同时除以2^(n+1)有A(n+1)/2^n+1=An/2^n+1/2设Bn=An/2^n则B(n+1)=Bn+0.5Bn是等差数列即An/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)+
我表示一楼很挫,楼主既然问这个问题不是找你要答案你总得写点过程吧an+1=an^2两边同时取对数lgan+1=2lgan则lgan为等比数列lgan=lga1*2^(n-1)an=a1^(2^(n-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
(1)a(n+1)/2^(n+1)=an/(an+2^n)2^(n+1)/a(n+1)=(an+2^n)/an=1+2^n/an2^(n+1)/a(n+1)-2^n/an=1所以{2^n/an}是以公
∵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+