已知数列{an}中,an=n•( 7 9 )n 1,此数列的最大项的项数是
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a(n+1)=2an/(an+1)∴1/a(n+1)=(an+1)/2an=1/2an+1/2∴1/a(n+1)-1=1/2an+1/2-1=1/2an-1/2=(1/2)(1/an-1),1/a1-
(1)数列{an}中,a1=1,前n项和Sn=n+23an,可知S2=43a2,得3(a1+a2)=4a2,解得a2=3a1=3,由S3=53a3,得3(a1+a2+a3)=5a3,解得a3=32(a
先求倒数1/a(n+1)=(an+2)/(2an)1/a(n+1)=1/2+(1/an)所以1/an是一个等差数列,公差d为1/2所以1/an=1/a1+(n-1)*d=1/a1+(n-1)/2
a(n+1)=an+lg[n/(n+1)]即a(n+1)-an=lgn-lg(n+1)将n=1,2,3,...代入,得a2-a1=lg1-lg2a3-a2=lg2-lg3.an-a(n-1)=lg(n
(Ⅰ)∵a1=-58,an+1-an=1n(n+1),∴a2=−18,a3=124  
因为an-2/an=2n所以:(an)^2-2nan-2=0根据万能公式:an=n-√(n^2+2),an=n+√(n^2+2)>0又因an<0所以:an=n-√(n^2+2),假设m>n>0那么am
a(n+1)=an^2+2ana(n+1)+1=(an+1)^2log2[(a(n+1)+1]=2log2[(an)+1]log2[(a(n+1)+1]/log2[an+1]=2{log2[a(n+1
∵数列{an}中,an=2n−1(n为正奇数)2n−1(n为正偶数),∴a9=29-1=28=256.S9=21-1+(2×2-1)+23-1+(2×4-1)+25-1+(2×6-1)+27-1+(2
n+1-bn=an+1-(n+1)^2+n+1-an+n^2-n等于一个常数,就可以证明是以神马为首项神马为公差的等比
A(n+2)=6*(n+1)*2^(n+1)-A(n+1)A(n+2)-A(n+1)=(6n+12)*2^n-A(n+1)+AnA(n+2)=(6n+12)*2^n+AnA3=37A2=11d=26A
由an+1=an+2n可以列出以下各式a10=a9+2x9a9=a8+2x8a8=a7+2x7..a3=a2+2x2a2=a1+2x1以上各式相加可得a10=a1+1x2+2x2+.+9x2a10=9
an+1-an=2^nan-an-1=2^n-1a2-a1=2^1-1an-a1=2^1+2^2+2^3+...2^n-1an=2^n+1
此类题目采用累加法或迭代法∵an+1-an=3n(往下递推)∴an-an-1=3(n-1)an-1-an-2=3(n-2).a3-a2=3×2a2-a1=3×1以上格式左边+左边=右边+右边左边相加的
a1=aa(n+1)+an=4n-1-->a(0+1)+a0=-1-->a1+a0=-1-->a0=-1-a(1)若a=1则a0=-1-1=-2a1=1a2=a(1+1)=4-1-a1=2a3=a(2
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
应该是A(n+1)=An+2n吧~~~=>a(n+1)-an=2n所以an-a(n-1)=2(n-1)a(n-1)-a(n-2)=2(n-2)...a2-a1=2*1把左边加起来,右边加起来得到an-
an=(n+1)(n+2)再问:有木有过程?再答:原式整理后得到an=(n+1)(an-1/n+1)试值:a2=(2+1)(6/2+1)=(2+1)(2x3/2+1)=12=3x4a3=(3+1)(1
A(n+1)=An+2(n+1)A(n+1)-An=2(n+1)即An-A(n-1)=2nA(n-1)-A(n-2)=2(n-1).A3-A2=2*3A2-A1=2*2以上各式相加得:An-A1=2*
An+1/An=[(n+2)(10/11)^n+1]/[(n+1)(10/11)^n]=[(n+2)/(n+1)]*(10/11)=(10n+20)/(11n+11)1.)当10n+20大于11n+1
sn/n=(2n-1)an(n>=1),sn=(2n^2-n)an,s(n+1)=(2n^2+3n+1)a(n+1),两者相减可得(2n+3)an+1=(2n-1)an,an=(2n-3)*a(n-1