in+an+instant

陆朗相3818in+the+first+instance是什么意思 -
汪砖印17878262979 ______ in the first instance.首先;起初;在初审时 The research team is also working towards making a STAIR cell prototype suited, in the first instance, for small applications, such as mobile phones or MP3 players. 研究团队还在努力制作合适的 STAIR 电池原型,起初用于一些小型电器上,比如移动电话和 MP3 播放器.

陆朗相3818对一组事件A1, A2, ...An ,必有P(A1+A2+ ...+An)=P(A1)+P(A2)+...
汪砖印17878262979 ______ 应该是:in answer to sth for sth 如:In answer to your inquiry for the price, our reply is as follows. 就贵方价格的询问,我们的答复如下.

陆朗相3818用构造法a1=1 An+1=3an+3n次方 -
汪砖印17878262979 ______[答案] an=3an-1+3^(n-1)代入可得an+1=3^2an-1+2*3^n an-1=3an-2+3^(n-2)代入 an+1=3^3an-2+3*3^n ... an+1=3^na1+n*3^n 即an=n*3^(n-1) (*为* ^几的几次方)

陆朗相3818已知数列{an}满足a1=1,an=Sn+(n+1)求an的递推公式 -
汪砖印17878262979 ______[答案] an+1=Sn+1+(n+1+1) an=Sn+(n+1) an+1-an=Sn+1+(n+1+1)-Sn+(n+1)=an+1+1 an=-1

陆朗相3818求An+1=An+n+1的通项公式 -
汪砖印17878262979 ______[答案] an+1=a1+2+3+4+……+n =a1+(2+n)(n-1)/2

陆朗相3818已知数列{an}中,a1=1,an+1=2an/(an+2)求数列{anan+1}的前n项和Sn -
汪砖印17878262979 ______[答案] 由已知可得1/(an+1)=1/an+1/2 易得1/an=(n+1)/2 所以an=2/(n+1) anan+1=2(1/n-1/(n+1)) 故Sn=2n/(n+1)

陆朗相3818项数为偶数2N的等差数列{an},证明:S2n=n(a1+a2n)=~=n(an+an+1)[an与an+1为中间两项】 -
汪砖印17878262979 ______[答案] 项数为偶数,所以都可以配对,共有N对 p,q,r,s为下标,当p+q=r+s时,有ap+aq=ar+as, 所以a1+a2n=a2+a2n-1=…=ak+a(2n-k+1)……=an+an+1,这n对的值都相等 所以S2n=n(a1+a2n)=……n(ak+a(2n-k+1)=……=n(an+an+1) 解毕

陆朗相3818a1=5分之1,an+a(n+1)=5n+1分之6,则an= -
汪砖印17878262979 ______[答案] an+a(n+1)=5n+1 an+an+a1=5n+1 2an=5n+1-a1 an=(5n+1-a1)*1/2 an=(5n+1-1/5)*1/2 an=(5n+1)/2+1/10 an=(25n+6)/10

陆朗相3818若a1>0,a(n+1)=1/2(an+1/an),an极限是否存在,若存在求之. -
汪砖印17878262979 ______[答案] 反证法: 假设极限存在,且等于A,则lim(n->∞)a(n+1)=lim(n->∞) 1/2(an+1/an) A=1/2(A+1/A) 2A^2+2=1 A^2=-1/2=0矛盾 所以极限不存在