早教吧作业答案频道 -->数学-->
数列高手请进设X(1)=1,X(n+1)=1+Xn/(X(n)+1),求X(n)的通项公式.
题目详情
数列高手请进
设X(1)=1,X(n+1)=1+Xn/( X(n)+1),求X(n)的通项公式.
设X(1)=1,X(n+1)=1+Xn/( X(n)+1),求X(n)的通项公式.
▼优质解答
答案和解析
∵x[n+1]=1+x[n]/(x[n]+1)=(2x[n]+1)/(x[n]+1)
∴不动点法,有:y=(2y+1)/(y+1),即:y^2-y-1=0
∴y=(1±√5)/2
∴[x[n+1]-(1+√5)/2]/[x[n+1]-(1-√5)/2]
=[(2x[n]+1)/(x[n]+1)-(1+√5)/2]/[(2x[n]+1)/(x[n]+1)-(1-√5)/2]
=[2(2x[n]+1)-(1+√5)(x[n]+1)]/[2(2x[n]+1)-(1-√5)(x[n]+1)]
=[4x[n]+2-x[n]-√5x[n]-1-√5]/[4x[n]+2-x[n]+√5x[n]-1+√5]
=[(3-√5)x[n]+1-√5]/[(3+√5)x[n]+1+√5]
=[(3-√5)/(3+√5)][x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=[(1-√5)/(1+√5)]^2[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
∵x[1]=1
∴[x[1]-(1+√5)/2]/[x[1]-(1-√5)/2]=(1-√5)/(1+√5)
∴{[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]}是:
首项为(1-√5)/(1+√5),公比为[(1-√5)/(1+√5)]^2的等比数列
即:[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=(1-√5)/(1+√5)[(1-√5)/(1+√5)]^[2(n-1)]
=[(1-√5)/(1+√5)]^(2n-1)
于是有:
x[n]-(1+√5)/2=x[n][(1-√5)/(1+√5)]^(2n-1)-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n]-x[n][(1-√5)/(1+√5)]^(2n-1)=(1+√5)/2-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n][(1+√5)^(2n-1)-(1-√5)^(2n-1)]=[(1+√5)^(2n)-(1-√5)^(2n)]/2
∴{x[n]}的通项公式:
x[n]=(1/2)[(1+√5)^(2n)-(1-√5)^(2n)]/[(1+√5)^(2n-1)-(1-√5)^(2n-1)]
∴不动点法,有:y=(2y+1)/(y+1),即:y^2-y-1=0
∴y=(1±√5)/2
∴[x[n+1]-(1+√5)/2]/[x[n+1]-(1-√5)/2]
=[(2x[n]+1)/(x[n]+1)-(1+√5)/2]/[(2x[n]+1)/(x[n]+1)-(1-√5)/2]
=[2(2x[n]+1)-(1+√5)(x[n]+1)]/[2(2x[n]+1)-(1-√5)(x[n]+1)]
=[4x[n]+2-x[n]-√5x[n]-1-√5]/[4x[n]+2-x[n]+√5x[n]-1+√5]
=[(3-√5)x[n]+1-√5]/[(3+√5)x[n]+1+√5]
=[(3-√5)/(3+√5)][x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=[(1-√5)/(1+√5)]^2[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
∵x[1]=1
∴[x[1]-(1+√5)/2]/[x[1]-(1-√5)/2]=(1-√5)/(1+√5)
∴{[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]}是:
首项为(1-√5)/(1+√5),公比为[(1-√5)/(1+√5)]^2的等比数列
即:[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=(1-√5)/(1+√5)[(1-√5)/(1+√5)]^[2(n-1)]
=[(1-√5)/(1+√5)]^(2n-1)
于是有:
x[n]-(1+√5)/2=x[n][(1-√5)/(1+√5)]^(2n-1)-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n]-x[n][(1-√5)/(1+√5)]^(2n-1)=(1+√5)/2-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n][(1+√5)^(2n-1)-(1-√5)^(2n-1)]=[(1+√5)^(2n)-(1-√5)^(2n)]/2
∴{x[n]}的通项公式:
x[n]=(1/2)[(1+√5)^(2n)-(1-√5)^(2n)]/[(1+√5)^(2n-1)-(1-√5)^(2n-1)]
看了 数列高手请进设X(1)=1,...的网友还看了以下:
数列{an}为正项等比数列,且满足;设正项数列{bn}的前n项和为Sn,满足.(1)求{an}的通 2020-05-13 …
各项均为正数的等比数列{an}满足a2=3,a4-2a3=9(1)求数列{an}的通项公式;(2) 2020-05-13 …
数列为正项等比数列,且满足;设正项数列的前n项和为Sn,满足.(1)求的通项公式;(2)设的前项的 2020-05-13 …
设、为实数,首项为,公差为的等差数列的前项和为,满足,.(1)求通项及;(2)设是首项为,公比为的 2020-05-14 …
设Sn是数列(An)的前n项和.点P(An Sn) 在直线y=2x-2上.(1)求数列(an)的通 2020-05-15 …
设数列{a}的前n项和为Sn,已知a1=a,an+1=Sn+3n次方设数列an的前n项和为Sn,已 2020-05-17 …
请教奥数问题,项数公式项数:(末项求和公式:(首项+末项)*项数/2项数:(末项-首项)/公差+1 2020-06-11 …
设多项式A是个三项式,B是个四项式,则A×B的结果的多项式的项数一定是()A.多于7项B.不多于7 2020-07-09 …
求一个类Polynimial多项式的函数设计多项式类Polynomial,多项式的每一项用数组表示 2020-07-27 …
高二数学二项式展开式项的系数最大值问题疑问求教,在学习二项式一章时,常见的一类题便是已知二项式,求 2020-08-03 …