已知等差数列an的公差不为0a1 =25,a3 a5=38
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即对任意n∈N,(a+n)/(a+n-1)≥(a+8)/(a+7)两边同减1:1/(a+n-1)≥1/(a+7)此不等式可分三种情况:(1)a+7≥a+n-1〉0显然n≥8时不成立(2)0〉a+n-1
a3²=a1a13(a1+2d)²+a1(a1+12d)a1=1所以1+4d+4d²=1+12d4d²-8d=0所以d=2所以an=2n-1bn=2^)2n-1
a2=a1+da4=a1+3da6=a1+5da2,a4-2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+
问题还没齐全吧?“求证;方程Akx6”什么意思呢?
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
(1)由题设可知公差d≠0,由a1=1且a1,a3,a9成等比数列,得:(1+2d)2=1+8d,解得d=1或d=0(舍去),故{an}的通项an=n.(2)∵bn=2 an=2n,∴数列{
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
1.设数列{an}的公差是d,则a(n+1)cosA+an*sinA=(an+d)*cosA+an*sinA=1即(cosA+sinA)*an=1-dcosA若cosA+sinA不等于0,则an=(1
再问:我本来也这么考虑的,但是如果a<1怎么办?再答:首先题目有没有与我理解得一样?再答:A小于l的情况分析起来很简单,但不可以小于零。再问:题目上说a属于R,麻烦你在考虑下。描述完整我就采纳你的答案
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+
a2=a1+da4=a1+3da6=a1+5da2,a4-2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+
设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
(1)设等差数列{an}的公差为d,由a22=a1a4,…(1分)得(a1+d)2=a1(a1+3d)…(2分)∵d≠0,∴d=a,∴an=na1,Sn=an(n+1)2.(2)∵1Sn=2a(1n−
a2^2=a1*a4,由等差得an=a1+(n-1)d,a1=a;带入得,an=na;则,1/a2+1/a2的平方+1/a2的3次+…+1/a2的n次与1/a1的大小等价于1/a2+1/a2的平方+1
a2,a5,a14是等比数列所以(a5)^2=a2*a14即(a+4d)^2=(a+d)*(a+13d)化简得d=2a所以公比q=a5/a2=(a+4*2a)/(a+2a)=3(2)a122=a+12
1/a1/(a+2d)=1/(a+d)^2a(a+2d)=(a+d)^2a^2+2ad=a^2+2ad+d^2d^2=0d=0哪儿写错了吧?再问:是1/a1,1/a2,1/a4成等比数列再答:a2^2
设an=a1+(n-1)d则a2=a1+da3=a1+2da4=a1+3da7=a1+6d因为等差数列{an}的前四项和为10所以,a1+a2+a3+a4=10即4a1+6d=10.①又因a2,a3,
{an}是首项为a公差为1的等差数列,∴数列{an}的通项公式为an=a+n-1,∵bn=1+anan=1+1an=1+1a+n−1.∵bn≥b8∴1+1an≥1+1a8,即1an≥1a8,数列{an