已知等差数列{an}和{bn}的前n项和分别为An和Bn,且An/Bn=(7n+41)/(n+3),
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/02 21:22:34
已知等差数列{an}和{bn}的前n项和分别为An和Bn,且An/Bn=(7n+41)/(n+3),
则使得an/bn为整数的正整数
则使得an/bn为整数的正整数
An/Bn=(7n+41)/(n+3)
=[n/2*(a1+an)]/[n/2*(b1+bn)]
=(a1+an)/(b1+bn)
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[2a1/d2+(n-1)d1/d2]/[2b1/d2+n-1]
=[d1/d2*n+2a1/d2-d1/d2]/(n+2b1/d2-1)
故有d1/d2=7
2a1/d2-d1/d2=41
2b1/d2-1=3
也即d1=7d2
a1=24d2
b1=2d2
于是an/bn=[a1+(n-1)d1]/[b1+(n-1)d2]
=[24d2+7(n-1)d2]/[2d2+(n-1)d2]
=(7n+17)/(n+1)
=(7n+7+10)/(n+1)
=7+10/(n+1)
故当n=1时,an/bn=12;
当n=4时,an/bn=9;
当n=9时,an/bn=8
也即只有n=1,4,9时,an/bn才为正整数.
=[n/2*(a1+an)]/[n/2*(b1+bn)]
=(a1+an)/(b1+bn)
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[2a1/d2+(n-1)d1/d2]/[2b1/d2+n-1]
=[d1/d2*n+2a1/d2-d1/d2]/(n+2b1/d2-1)
故有d1/d2=7
2a1/d2-d1/d2=41
2b1/d2-1=3
也即d1=7d2
a1=24d2
b1=2d2
于是an/bn=[a1+(n-1)d1]/[b1+(n-1)d2]
=[24d2+7(n-1)d2]/[2d2+(n-1)d2]
=(7n+17)/(n+1)
=(7n+7+10)/(n+1)
=7+10/(n+1)
故当n=1时,an/bn=12;
当n=4时,an/bn=9;
当n=9时,an/bn=8
也即只有n=1,4,9时,an/bn才为正整数.
等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn
已知{an},{bn}均为等差数列,前n项的和为An,Bn,且An/Bn=2n/(3n+1),求a10/b10的值
已知两个等差数列{an}和{bn}的前n项和分别为An,Bn,且An/Bn=(3n-3)/(2n+3),则a6/b6=
已知两个等差数列{an}和{bn}的前n项和分别为An和Bn,且A
已知等差数列{an}和{bn}前n项和为An和Bn,且An/Bn为7n+45/n+3,则使得an/bn为整数的n有几个,
已知等差数列an和bn的前n项和分别为An,Bn,且An/Bn=(3n+1)/(2n-1).若ak/bk=34/21,则
已知等差数列{an}、{bn}的前n项和分别为Sn、Tn,若Sn/Tn=【7n+1】/【4n+27】,则an/bn=
已知数列an满足bn=an-3n,且bn为等比数列,求an前n项和Sn
关于数列和 不等式.1.若两等差数列{an}{bn}的前n项和为 An Bn ,满足(An/Bn)=(7n+1)/4n+
1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=
已知两个等差数列an和bn的前n项和分别为Sn,Tn ( 1)若Sn/Tn=(7n+2)/(n+3) 求an/bn 2)
(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/