已知数列为等差数列,若a2 a3 a4 a5=34,a2a5=52

在原式基础上,再写一相同结构等式,到an+2结束.减去原式便得到：1/（an+1）an=n+1/（an+1）（an+2）-n/anan+1整理得…你题目可能出错了,不是等差数列.我们假设公差为d.那么

a(n+1)=2(n+1)-1所以an=2n-1所以Sn=1/1*3+1/3*5+……+1/197*199=(1/2)(1-1/3)+(1/2)(1/3-1/5)+……+(1/2)(1/197-1/1

1/a1a2+1/a2a3+…+1/anan+1=[(a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1]/d=[1/a1-1/a2+1/a2-1/a3+.

sn=1/d(1/a1-1/a2+1/a2-1/a3+.+1/an-1/a(n+1))=1/d(1/a1-1/a(n+1))=nd/da1a(n+1)=n/a1a(n+1)

n=3时1/a1a2+1/a2a3=2/a1a3两边乘以a1a2a3得到a3+a1=2a2前三项满足等差数列当n>=3时1/a1a2+1/a2a3+……1/an-1an=(n-1)/a1an①1/a1

因为1/a1a2+1/a2a3=2/a1a3,所以a3+a1=2a2即a2-a1=a3-a2所以a1,a2,a3成等差数列再问：a3+a1=2a2，这是为什么啊。再答：等式两边同乘以a1a2a3,即得

这是裂相求和.原式=1/a1-1/a2008.a2008=a12007d=12007x2=4015所以原式=4014/4015

因为1/anan+1=1/an*（an+d）=1/d[1/an-1/（an+d）]=1/d[1/an-1/an+1]所以1/a1a2+1/a2a3+…+1/anan+1=1/d[1/a1-1/a2+1

1/a1=1d=2所以1/an=(2n-1)所以原式=1/1*3+1/3*5+……+1/(2n-1)(2n+1)=(1/2)(1-1/3)+(1/2)(1/3-1/5)+……+(1/2)[1/(2n-

(a2)(a3)=45(a1)＋(a4)=(a2)＋(a3)=14解得：a1=5,a3=9则：d=4从而有：an=4n＋1Sn=[n(a1＋an)]/2=2n²＋3n则：bn=Sn/(n＋c

原式=1/(5×7)+1/(7×9)+1/(9×11)+.+1/[(2n+3)(2n+5)]=1/2[(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+.+1/(2n+3)-1/(2n+5

a2a3=45a1＋a4=a2＋a3=14解得：a2=5,a3=9则：d=a3-a2=4从而有：an=4n-3a1=1Sn=[n(a1＋an)]/2=2n²-n

an=an+1（1+2an）an/（1+2an）=an+11/an+1=1/an+21/an=1+(n-1)2=2n-1an=1/(2n-1)2anan+1=2/(2n-1)*1/(2n+1)=1/(

恳请题目能出全点,否则我怎么做?要求助我哦~我帮你!

1.设a1=a则1/a1a2+1/a2a3+.+1/a(n-1)an=1/a(a+d)+1/(a+d)(a+2d)+……+1/[a+(n-2)d][a+(n-1)d]={d/a(a+d)+d/(a+d

inta[3]={1,1,2};inttemp;for(inti=3;i再问：inta[3]={1,1,2};inttemp;是什么意思啊？求解再答：存前面3个数啊

设公差为d,d不等于01/a1a2+1/a2a3+1/a3a4+...+1/an-1an=(1/d)*(1/a1-1/a2+/1a2-1/a3+1/a3-1/a4.+1/an-1-1/an)=1/d(

2*s2=s3+s4带入公式2*a1*(1-q^2)/(1-q)=a1*(2-q^3-q^4)/(1-q)得q^2=q^3+q^4消去q^2q^2+q-2=0得q=1或q=-2但q不能为1则有an=4

证明：左边＝1/(a1a2)+1/(a2a3)+...+1/(an-1*an)＝1/d(1/a1-1/a2)+1/d(1/a2-1/a3)+...+1/d(1/an-1-1/an)=1/d[(1/a2

a2=a1+da3=a1+2da4=a1+3da2a3=(a1+d)(a1+2d)=45a1+a4=a1+a1+3d=14a1=1d=4an=a1+4(n-1)