已知正数数列{an}满足2根号Sn=an 1,(1)计算a1,a2,a3,a4

a(n+1)=(1/2)an(4-an)2a(n+1)=4an-an^2=-[an^2-2*2an+4]+4=-(an-2)^2+42[a(n+1)-2]=-(an-2)^2设bn=an-2,b0=a

∵Sn-Sn-1=√Sn+√Sn-1∴(√Sn)²-(√Sn-1)²=√Sn+√Sn-1(√Sn-√Sn-1)(√Sn+√Sn-1)=√Sn+√Sn-1∴√Sn-√Sn-1=1（n

(1)当n=0时,显然成立(2)假设当n=k时,ak

an+1-2=-1/2(an_2)^2所以an-2=-1/2(an-1-2)^2=-1/2^(1+2)(an-2-2)^4=...=-1/2^(2^0+2^1+.2^(n-1))(a0-2)^(2^n

2倍的根号下Sn=An+1根号下Sn=(An+1)/2Sn=(An+1)^2/4An=Sn-S(n-1)=(An+1)^2/4-(A(n-1)+1)^2/4即：4An=(An)^2+2An-[A(n-

an+Sn=4a(n-1)+S(n-1)=4相减：an/a(n-1)=1/2等比数列n=1时a1+a1=4a1=2an=2^(2-n)bn=1/n²数学归纳法n=2时T2=5/4

2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1

设bn=根号an所以A(n-1)-An=(2倍根号An)+1等于根号[b(n-1)]^2-bn^2=2bn+1即[b(n-1)]^2=(bn+1）^2因为{a}中各项为正数,且a1=2所以b(n-1)

∵（an+1）²-an+1×an-2an²=0∴（an+1+an)(an+1-2an)=0∴an+1-2an=0,an+1+an=0(舍去）∴an+1=2an∴an是等比数列,设a

1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=

你看下2009年广东文科数学高考试题中的第20题,该题应该是由它变化过来的,只是改了一点点.你仿照一下它的解法就行了!

由题意得an^2+2根号n*an-1=0解出来以后讨论下,因为an>0an=-根号下n+根号下n+1

（1）(Xn)^an=(Xn+1)^an+1=(Xn+2)an+2=k得Xn=k^(1/an),X(n+1)=k^(1/a(n+1)),X(n+2)=k^(1/(an+2))由等比数列｛Xn｝可知：(

6Sn=an^2+3an+26S(n-1)=a(n-1)^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-a(n-1)^2-3a(n-1)-26an=an^2+3an-a(

a1=2>0假设当n=k(k∈N+)时,ak>0,则a(k+1)=3√ak>0k为任意正整数,因此对于任意正整数n,an恒>0,数列各项均为正.a(n+1)=3√anlog3[a(n+1)]=log3

因为不清楚你写的到底是怎样,我把我理解出的可能的两种题目都写出来.①假定原题为1/(An+1)=√[1/(An²+2)]两边同时平方,有1/(An+1)²=1/(An²+

6Sn=An^2+3An+26S(n-1)=[A(n-1)]^2+3A(n-1)+26Sn-6S(n-1)=6An=An^2+3An+2-{[A(n-1)]^2+3A(n-1)+2}An-A(n-1)

1.A(n+1)^2*An+A(n+1)*An^2+A(n+1)^2-An^2=0两边同除以A(n+1)²An²1/An+1/A(n+1)+1/An²-1/A(n+1)&

本题需要先对an的取值范围进行判断,然后才能用取对数、用待定系数法,因此过程比较复杂.a0=10

a(3)=a(1+2)=1/[1+a(1)]=a(1),1=a(1)+[a(1)]^2,0=[a(1)]^2+a(1)-1,Delta=1+4=5.a(1)=[-1+5^(1/2)]/2,或a(1)=