Sn =n^2 (1)
S(n-1) = (n-1)^2 (2)
(1) -(2)
an = 2n-1
consider
1/(ana(n+1))
=1/[(2n-1)(2n+1)]
=(1/2)[ 1/(2n-1) -1/(2n+1)]
Tn=1/(a1a2)+1/(a2a3) +..+1/(ana(n+1))
=(1/2)[ (1-1/3) + (1/3-1/5) +...+(1/(2n-1)-1/(2n+1) ]
= (1/2)(1- 1/(2n+1)]
= n/(2n+1)
Sn =n^2 S(n-1) = (n-1)^2 (1) -(2)
an = 2n-1
consider
1/(ana(n+1))
=1/[(2n-1)(2n+1)]
=(1/2)[ 1/(2n-1) -1/(2n+1)]
Tn = n/(2n+1)
已知数列{an}的前n项和Sn=n^2.求Tn=1\/a1a2+1\/a2a3+……+1\/ana(n+1)
Sn =n^2 (1)S(n-1) = (n-1)^2 (2)(1) -(2)an = 2n-1 consider 1\/(ana(n+1))=1\/[(2n-1)(2n+1)]=(1\/2)[ 1\/(2n-1) -1\/(2n+1)]Tn=1\/(a1a2)+1\/(a2a3) +..+1\/(ana(n+1))=(1\/2)[ (1-1\/3) + (1\/3-1\/5) +...+(1\/(2n-1)-1\/(2n...
...前n项和Sn=n^2,记Pn=1\/(a1*a2)+1\/(a2*a3)+...+1\/(an*an+1),求Pn...
==> Pn=[1\/a1-1\/a2 + 1\/a2-1\/a3 + ……+ 1\/an-1\/a(n+1)]\/4 =[1\/a1 - 1\/a(n+1)]\/4 =[1-1\/(2n+1)]\/4 所以极限是 1\/4 注：n趋于无穷时,1\/(2n+1)趋于0 回答完毕
...a N】的前N项和Sn=n2+2。设T n=1\/a1a2+1\/a2a3+~~~+1\/anan+1,求...
二楼错了，应该是 n>=2时 Sn = n² + 2 S(n+1) = n² + 2n + 3 a(n+1) = S(n+1) - Sn = 2n + 1 则an = 2n -1 a1=S1=3 Tn = 1\/(3*3) + 1\/(3*5)+1\/(5*7) + ... + 1\/[(2n-1)*(2n+1)]= 1\/9+1\/2 * [1\/3 - 1\/5 + 1\/5 ...
设数列{an}的前n项和Sn=n2,如果Pn=1a1a2+1a2a3+…+1anan+1,则limn→...
∵Sn=n2，∴a1=s1=1，n≥2时，an=sn-sn+1=n2-（n-1）2=2n-1，对n=1时也成立，∴an=2n-1．∴1anan+1=1(2n?1)(2n+1)=12（12n?1-12n+1），∴Pn=1a1a2+1a2a3+…+1anan+1=12（1?13+13?15+…+12n?1-12n+1）=12（1-12n+1）=12-14n+2，∴limn→∞Pn=limn→...
已知数列{an}的前n项和为Sn=1\/2n2+1\/2n(n?n*)(1)求数列{an}的通项公...
an=Sn-Sn-1=[1\/2n2+1\/2n]-[1\/2(n-1)^2+1\/2(n-1)]=n an=n Sn=1\/2n2+1\/2n 1\/Sn=2\/n(n+1)=2[1\/n-1\/n+1](2)记T=1\/s1+1\/s2+1\/s3+…+1\/s99=2[1-1\/2+1\/2-\/13+……+1\/99-1\/100]=2*(1-1\/100)=99\/50 ...
已知数列{an}前n项和Sn=n^2+2n (1)求数列的通项公式an (2)设Tn=1\/...
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已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足bn=2\/(an)+1,前n项和...
你好：(1)∵数列{an}满足前N项和sn=n平方+1 ∴Sn=n^2+1 S(n-1)=(n-1)^2+1 An=Sn-S(n-1)=n^2+1-[(n-1)^2+1]=2n-1 A1=S1=2 Bn=2\/An +1=2\/(2n-1)+1=(2n+1)\/(2n-1)B1=2\/A1+1=2 Bn是一个首项为2，通项为(2n+1)\/(2n-1) 的数列 (2)Cn=T(2n+...
已知数列{an}前n项和Sn=n^2+n,令bn=1\/anan+1,求数列{bn}的前n项和Tn
解：n＝1,S1=a1=2 , n>1 , an=Sn－S(n-1)＝2n, n=1时也适合，故：an=2n bn=(1\/4)·1\/n(n+1) 4bn=1\/n(n+1) =1\/n－1\/(n+1),所以：4Tn=[(1－1\/2)+(1\/2－1\/3)+···+(1\/n－1\/(n+1))] ＝1－1\/(n＋1)＝n\/(n+1)Tn＝n\/4(n+1)
已知数列{an}的前n项和为Sn=n2+n求数列{an}的通项公式;若bn=(1\/2...
解：(1)a1=S1=1^2+1=2 Sn=n^2+n Sn-1=(n-1)^2+(n-1)an=Sn-Sn-1=n^2+n-(n-1)^2-(n-1)=2n {an}通项公式为an=2n (2)bn=(1\/2)^an+n=(1\/2)^(2n)+n=(1\/4)^n+n Tn=b1+b2+...+bn =(1\/4)^1+(1\/4)^2+...+(1\/4)^n+(1+2+...+n)=(1\/...
已知数列{an}的前n项和为Sn,且a1=1\/2,a(n+1)=(n+1)an\/2n,(1)求{an}...
a1\/1=(1\/2)\/1=1\/2，数列{an\/n}是以1\/2为首项，1\/2为公比的等比数列 an\/n=(1\/2)(1\/2)^(n-1)=1\/2ⁿan=n\/2ⁿ数列{an}的通项公式为an=n\/2ⁿ(2)Sn=a1+a2+a3+...+an=1\/2+2\/2²+3\/2³+...+n\/2ⁿSn \/2=1\/2²+2\/2...
