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《常见递推数列通项公式的求法典型例题及习题》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、常见递推数列通项公式的求法典型例题及习题【典型例题】a=ka+b[例1]n+1n型。a-a=bÞ{a}a=b×n+(a-b)(1)k=1时,n+1nn是等差数列,n1a+m=k(a+m)a=ka+km-m(2)k¹1时,设n+1n∴n+1nbm=比较系数:km-m=b∴k-1bb{a+}a+n1∴k-1是等比数列,公比为k,首项为k-1bbn-1bn-1ba+=(a+)×ka=(a+)×k-n1n1∴k-1k-1∴k-1k-1a=ka+f(n)[例2]n+1n型。a-a=f(n)(1)k=1时,n+1n,若f(n)可求和,则
2、可用累加消项的方法。1a-a=n+1n例:已知{an}满足a1=1,n(n+1)求{an}的通项公式。解:111a-a==-n+1n∵n(n+1)nn+11111a-a=-a-a=-nn-1n-1n-2∴n-1nn-2n-111a-a=-n-2n-3n-3n-2……111a-a=-a-a=1-322123211a-a=1-a=2-n1n对这(n-1)个式子求和得:n∴na+A(n+1)+B=k(a+An+B)(2)k¹1时,当f(n)=an+b则可设n+1na=ka+(k-1)An+(k-1)B-A∴n+1nì(k-1)A=
3、aabaíA=B=+2∴î(k-1)B-A=b解得:k-1,k-1(k-1){a+An+B}∴n是以a1+A+B为首项,k为公比的等比数列n-1a+An+B=(a+A+B)×k∴n1n-1a=(a+A+B)×k-An-B∴n1将A、B代入即可n(3)f(n)=q(q¹0,1)an+1kan1=×+n+1n+1nqqqqq等式两边同时除以得ank1Cn=nCn+1=Cn+令q则qq∴{Cn}可归为an+1=kan+b型a=f(n)×a[例3]n+1n型。(1)若f(n)是常数时,可归为等比数列。(2)若f(n)可求积,可用累积
4、约项的方法化简求通项。12n-1a=a=a1nn-1{a}例:已知:3,2n+1(n³2)求数列n的通项。anan-1an-2a3a22n-12n-32n-5533××L×=××L×=aaaaa2n+12n-12n-3752n+1解:n-1n-2n-32131a=a×=n1∴2n+12n+1m×an-1a=k×nm+a[例4]n-1型。11111k=k(+)=k×+aamaam考虑函数倒数关系有nn-1∴nn-11C=na{C}a=ka+b令n则n可归为n+1n型。练习:{a}a=2a+11.已知n满足a1=3,n+1n求通
5、项公式。解:a+m=2(a+m)a=2a+m设n+1nn+1n∴m=1{a+1}∴n+1是以4为首项,2为公比为等比数列n-1n+1a+1=4×2a=2-1∴n∴n{a}a=a+2n*2.已知n的首项a1=1,n+1n(nÎN)求通项公式。解:a-a=2(n-1)nn-1a-a=2(n-2)n-1n-2a-a=2(n-3)n-2n-3……a-a=2´232+a-a=2´1212a-a=2[1+2+L+(n-1)]=n-nn12a=n-n-1∴nna=a{a}n+1n3.已知n中,n+2且a1=2求数列通项公式。解:anan-
6、1an-2a3a2n-1n-2n-3n-4212××L×=×××L×=aaaaan+1nn-1n-243n(n+1)n-1n-2n-321an24=a=n∴a1n(n+1)∴n(n+1)n+12×ana=n+1n+14.数列{an}中,2+an,a1=2,求{an}的通项。解:n+112+an111==+n+1n+1an+12an∴an+1an2111b=nb=b+b=b+an+1n2n+1nn-12n设n∴∴1b-b=nn-1n∴21b-b=n-1n-2n-121b-b=n-2n-3n-22……1b-b=32321+b-b
7、=212211n-1[1-()]22211==-111122nb-b=++L+1-n123n2222nn1112-12b=-+=a=nnnnn∴2222∴2-11a=a+2n-1nn-1{a}5.已知:a1=1,n³2时,2,求n的通项公式。解:1a+An+B=[a+A(n-1)+B]nn-1设21111a=a-An-A-Bnn-12222ì1-A=2ïï2í11ìA=-4ï-A-B=-1í∴ïî22解得:îB=6∴a1-4+6=31{a-4n+6}∴n是以3为首项,2为公比的等比数列1n-13a-4n+6=3×()a=+4
8、n-6nnn-1∴2∴2【模拟试题】n{a}a=a+2a1.已知n中,a1=3,n+1n,求n。{a}a=3a+2a2.已知n中,a1=1,nn-1(n³2)求n。n{a}a=2a+2a3.已知n中,a1=1,nn-1(n³2)求n。4a=4-n{a}aa4.已知n中,a1=4,n-1(n
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