数列an=4n-1中删掉被3或5整除
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第1问:设数列{bn},令bn=an-n则an=bn+n代入a(n+1)=4an-3n+1得b(n+1)+n+1=4(bn+n)-3n+1化简得b(n+1)=4bn所以数列{bn}即数列{an-n}是
1.a(n+1)=4an-3n+1=>a(n+1)-(n+1)=4(an-n){an-n}是等比数列2.an-n=4^(n-1)*(a1-1)=4^(n-1)=>an=4^(n-1)+nSn=(1+4
1.a(n+1)=4an-3n+1=4(an-n)+(n+1)a(n+1)-(n+1)=4[an-n]故:{an-n}是以a1-1=1为首项,4为公比的等比数列(2)由(1)得:an-n=4^(n-1
由条件得a1=2,a2=5.且有:a2-a1=3*1,a3-a2=3*2,a4-a3=3*3,...an-a(n-1)=3*(n-1),累加得,an-a1=3*(1+2+3+...+n-1)=3n(n
设:(An+1)+p(n+1)+q=4[An+pn+q]解得p=-1,q=0即An+1=4An-3n+1等价于(An+1)-(n+1)=4(An-n)若设Bn=An-n则Bn+1=4Bn则Bn=B1*
a(n+1)=4an-3n+1⇒a(n+1)-(n+1)=4(an-n)所以{an-n}是以4为公比的等比数列且a1-1=2-1=1an-n=4^(n-1)an=4^(n-1)+n
sn=3*3^1+5*3^2+.+(2n+1)*3^n①3sn=3*3^2+5*3^3+.+(2n-1)*3^n+(2n+1)*3^(n+1)②①-②-2Sn=Sn-3Sn=-2n*3^(n+1),因
1.an=-a(n-1)-2n+1an+n=-a(n-1)-n+1=-[a(n-1)+(n-1)](an+n)/[a(n-1)+(n-1)]=-1,为定值.a1+1=3+1=4数列{an+n}是以4为
(1)由an+1=4an-3n+1得[a(n+1)-(n+1)]/(an-n)=4所以数列{an-n}是公比为4的等比数列(2)设数列{an-n}的通项为bn,前n项的和为Tnb1=a1-1=1Tn=
(1)∵an+1=4an-3n+1n∈N*,∴an+1-(n+1)=4an-3n+1-(n+1)…(4)分=4an-4n=4(an-n)…(6)分∴{an-n}为首项a1-1=1,公比q=4的等比数列
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
a(n+1)=4an-3n+1a(n+1)-(n+1)=4an-3n+1-(n+1)a(n+1)-(n+1)=4(an-n)[a(n+1)-(n+1)]/[(an-n)]=4数列a(n)-n是公比为4
a(n+1)=-an+3n-54a(n+1)+x(n+1)+y=-an+3n-54+x(n+1)+ya(n+1)+x(n+1)+y=-[an-(3+x)n+54-x-y]令x=-(3+x)y=54-x
(1)a(n+1)=4an-3n+1a(n+1)-(n+1)=4an-3n+1-(n+1)a(n+1)-(n+1)=4(an-n)[a(n+1)-(n+1)]/[(an-n)]=4数列a(n)-n是公
(Ⅰ)证明:由题设an+1=4an-3n+1,得an+1-(n+1)=4(an-n),n∈N*.又a1-1=1,所以数列{an-n}是首项为1,且公比为4的等比数列.(Ⅱ)由(Ⅰ)可知an-n=4n-
(I)∵an+1=3an-2an-1(n≥2)∴(an+1-an)=2(an-an-1)(n≥2)∵a1=2,a2=4∴a2-a1=2≠0,∴an+1-an≠0故数列{an+1-an}是公比为2的等比
an+1=4an-3n+1an+1-(n+1)=4[an-n][an+1-(n+1)]/[an-n]=4等比a1-1=3an-n=3*4^(n-1)an=3*4^(n-1)+n2\sn=[3*4^(n
(1)∵an+1+an=3n−54an+2+an+1=3n−51,两式相减得an+2-an=3,∴a1,a3,a5,…,与a2,a4,a6,…都是d=3的等差数列∵a1=-20∴a2=-31,①当n为
a(n+1)-(n+1)=4an-3n+1-(n+1)=4an-4n得证
An+1=4An-3n+1An+1-(n+1)=4An-4nAn+1-(n+1)=4(An-n)[An+1-(n+1)]/[(An-n)]=4即:An-n是等比数列An-n=4^(n-1)An=4^(