1/2！+2/3！+3/4！+……+n/（n+1）！求和怎么求啊？ 在线等！_百度知道
1/2！+2/3！+3/4！+……+n/（n+1）！求和怎么求啊？ 在线等！
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..！+1/n;2;n！-1&#47！=1/3;（n+1);2利用n/2;2;4。!=1-1&#47!-1/（n+1）！=1&#47！-1&#47！ ;3！=1&#47.;1！+2/2;1！+;n;(n+1);(n+1);3.!+1&#47！
2&#47.！+……+n/2.+1&#47！+3&#47!-1&#47。！-1/3!得。1/(n-1);(n+1)：1&#47!-1&#47!=1&#47！-1&#47
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出门在外也不愁｛an}前n项和为Sn,Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn_作业帮
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｛an}前n项和为Sn,Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn
｛an}前n项和为Sn,Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn
当n=1时,a1=S1=(4/3)a1-(1/3)*2^(1+1)+2/3=(4/3)a1-2/3,解得:a1=2；当n>1时：Sn=(4/3)an-(1/3)*2^(n+1)+2/3=(4/3)an-2*(1/3)*2^n+2/3S(n-1)=(4/3)a(n-1)-(1/3)*2^n+2/3=(4/3)a(n-1)-1*(1/3)*2^n+2/3an=Sn-S(n-1)=[(4/3)an-2*(1/3)*2^n+2/3]-[(4/3)a(n-1)-1*(1/3)*2^n+2/3]=(4/3)an-(4/3)a(n-1)-(1/3)*2^n∴(1/3)an=(4/3)a(n-1)+(1/3)*2^n即 an=4*a(n-1)+2^n4*a(n-1)=4^2*a(n-2)+4*2^(n-1)……4^(n-2)*a2=4^(n-1)*a1+4^(n-2)*2^2上述式子相加,得：an=4^(n-1)*a1+2^n+4*2^(n-1)+…+4^(n-2)*2^2=2^(2n-2)*2+2^n+2^2*2^(n-1)+…+2^(2n-4)*2^2=2^(2n-1)+2^n+2^(n+1)+…+2^(2n-2)=2^(2n-1)+2^n[2^0+2^1+…+2^(n-2)]=2^(2n-1)+2^n*2^0*[1-2^(n-1)]/(1-2)=2^(2n-1)+2^n*[2^(n-1)-1]=2^(2n-1)+2^(2n-1)-2^n=2^1*2^(2n-1)-2^n=2^(2n)-2^n∵a1=2=2^2-2^1,符合上式∴数列{an}的通项公式是an=2^(2n)-2^n.(2)证明：Sn=(2^2-2^1)+(2^4-2^2)+…+[2^(2n)-2^n]=[2^2+2^4+…+2^(2n)]-(2^1+2^2+…+2^n)=4[1-(2^2)^n]/(1-2^2)-2(1-2^n)/(1-2)=(4/3)[(2^n)^2-1]-2(2^n-1)=(4/3)*(2^n)^2-4/3-2*2^n+2=(4/3)*(2^n)^2-2*2^n+2/3则Tn=2^n/Sn=1/[(4/3)*(2^n)-2+2/(3*2^n)]=(3/2)*1/(2*2^n+1/2^n-3).设f(n)=1/(2*2^n+1/2^n-3)=(2^n)/[2*(2^n)^2+1-3*(2^n)]=(2^n)/(2^n-1)(2*2^n-1)=[(2*2^n-1)-(2^n-1)]/(2^n-1)(2*2^n-1)=1/(2^n-1)-1/[2^(n+1)-1]则Tn=(3/2)*f(n)=(3/2)*{1/(2^n-1)-1/[2^(n+1)-1]}.T1+T2+T3+…+Tn=(3/2)*{(1-1/3)+(1/3-1/7)+(1/7-1/15)+…1/(2^n-1)-1/[2^(n+1)-1]}=(3/2)*{1-1/[2^(n+1)-1]}=3/2-(3/2)*{1/[2^(n+1)-1]}
(1) Sn=4/3an-1/3*2^（n+1）+2/3，S(n+1)=4/3a(n+1)-1/3*2^（n+2）+2/3，两式相减得a(n+1)=4an+2^(n+1)，两边同除以2^(n+1) 得 a(n+1)/ 2^(n+1)=2an/2^n+1令bn= an/2^n，得 b(n+1)=2bn+1，两边同加1得b(n+1)+1=2(bn+1)，∴﹛bn+1﹜是首项为b1+...& 2013 - 2014 作业宝. All Rights Reserved. 沪ICP备号-91\1*2*3+1\2*3*4+1\3*4*5+.+1\n(n+1)*(n+2)=_作业帮
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1\1*2*3+1\2*3*4+1\3*4*5+.+1\n(n+1)*(n+2)=
1\1*2*3+1\2*3*4+1\3*4*5+.+1\n(n+1)*(n+2)=
首先,1/（1*2*3）分成1/1*3 -1/2*3 1/（2*3*4）分成1/2*4 -1/3*4后面的一次类推,再分一下组,相加的一组相减的一组.我们先算相减的一组是 -1/2+1/3-1/3+1/4-1/4+1/5……-1/n+1+1/n+2得出的结果是 -1/2+1/101下面算相加的一组,相加的一组要先乘以2,最后再除以2,我们算乘2后,可以变为1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+{1/（n-2)}-{1/n}+{1/（n-1)}-{1/n+1}+{1/n}-{1/n+2}这样最后还剩余的项有1+1/2-{1/n+1}-{1/n+2}这四项,因为我们乘过2,所以现在再除以2,得到1/2+1/4-{1/2(n+1)}-{1/2(n+2)}最后把剩余的项加起来1/2+1/4-{1/2(n+1)}-{1/2(n+2)}-1/2+1/n+2得到最后的结果是1/4-1/2(n+1)+1/2(n+2)
通式化简1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)]=1/2[1/n-2/(n+1)+1/(n+2)]；代入通式所以，原式=1/2[(1/1-2/2+1/3)+(1/2-2/3+1/4)+(1/3-2/4+1/5)+……1/n-2/(n+1)+1/(n+2)]=1/2[1/1-2/2+1/2+1/(n+1)-2/(n+1)+1/(n...｛an}前n项和为Sn，Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn&3/2_百度知道
｛an}前n项和为Sn，Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn&3/2
详细过程。谢谢！！
提问者采纳
3)an-2*(1/2)*{1/3-1/3)*2^(1+1)+2/3)a1-2/3&#47，符合上式∴数列{an}的通项公式是an=2^(2n)-2^n;[2^(n+1)-1]则Tn=(3/2)*{1&#47：Sn=(4&#47:a1=2;2)*f(n)=(3/3)*2^n+2&#47；当n&3)a(n-1)-(1/(2^n-1)-1/3)*2^n∴(1/3)*2^n+2/[2*(2^n)^2+1-3*(2^n)]
=(2^n)/(2^n-1)(2*2^n-1)
=1/3)*(2^n)^2-2*2^n+2/3=(4/7)+(1/(1-2)
=2^(2n-1)+2^n*[2^(n-1)-1]
=2^(2n-1)+2^(2n-1)-2^n
=2^1*2^(2n-1)-2^n
=2^(2n)-2^n∵a1=2=2^2-2^1;3)*2^n即
an=4*a(n-1)+2^n
4*a(n-1)=4^2*a(n-2)+4*2^(n-1)
……4^(n-2)*a2=4^(n-1)*a1+4^(n-2)*2^2上述式子相加：Sn=(2^2-2^1)+(2^4-2^2)+…+[2^(2n)-2^n]
=[2^2+2^4+…+2^(2n)]-(2^1+2^2+…+2^n)
=4[1-(2^2)^n]/(2^n-1)-1/3;[2^(n+1)-1]}
=3/1时;3则Tn=2^n/3S(n-1)=(4&#47.设f(n)=1/(1-2^2)-2(1-2^n)/2^n-3);2)*{1-1/(2^n-1)(2*2^n-1)
=[(2*2^n-1)-(2^n-1)]/3)*2^n+2/3)a(n-1)-(1/15)+…1/3)an-(1/3)*(2^n)-2+2/3-2*2^n+2
=(4/(2*2^n+1/3)*2^(n+1)+2/2-(3/3)*(2^n)^2-4/3=(4/3]=(4/(2*2^n+1/2^n-3)
=(2^n)&#47，得;3=(4/2)*1/3)+(1&#47.(2)证明;(2^n-1)-1&#47，解得;3)an-(4/3)a(n-1)-1*(1/(1-2)
=(4/3]-[(4/3)a(n-1)-1*(1/3)an-2*(1/3 an=Sn-S(n-1)=[(4/3)*2^n+2&#47.T1+T2+T3+…+Tn=(3/3)[(2^n)^2-1]-2(2^n-1)
=(4/3)an=(4/[2^(n+1)-1]};7-1/[2^(n+1)-1]}
=(3&#47：当n=1时：an=4^(n-1)*a1+2^n+4*2^(n-1)+…+4^(n-2)*2^2
=2^(2n-2)*2+2^n+2^2*2^(n-1)+…+2^(2n-4)*2^2
=2^(2n-1)+2^n+2^(n+1)+…+2^(2n-2)
=2^(2n-1)+2^n[2^0+2^1+…+2^(n-2)]
=2^(2n-1)+2^n*2^0*[1-2^(n-1)]/Sn=1/3)a(n-1)+(1/(3*2^n)]=(3/[(4&#47，a1=S1=(4&#47(1)解;3)a1-(1/[2^(n+1)-1]}
&2)*{(1-1/3)*2^n+2&#47
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谢谢！！！
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(2^n-1)－1&#47：(1) Sn=4/3－1/(2*2^n-1)]故Tn=3/3，b1=a1/3+1/3*2^（n+1）+2&#47，即a1=2;3=2/3得a1=4&#47解;7＋1&#47，两边同除以2^(n+1) 得 a(n+1)/2[1－1/3*2^2n -2*2^n+2/(2*2^n-1)] &3得Sn=4&#47，得 b(n+1)=2bn+1，由Sn=4/3;3*2^(1+1)+2/3*2^（n+2）+2/3an-1&#47，公比为2的等比数列;2^n;2[1&#47，∴bn+1=( b1+1)* 2^(n-1)= 2^n∴bn= 2^n-1;3*2^（n+1）+2/3&#47，S(n+1)=4/3;Sn=3*2^n&#47，∴Tn=2^n/(2*2^n-1)]
=3/2=1;3a(n+1)-1/2^n+1令bn= an/2(2^n-1)( 2*2^n-1)=3/3a1-1/7－……＋1&#47，两式相减得a(n+1)=4an+2^(n+1)，∴﹛bn+1﹜是首项为b1+1;(2^n-1)－1/ 2^(n+1)=2an/3an-1/3(2^n-1)( 2*2^n-1);2[1－1&#47，两边同加1得b(n+1)+1=2(bn+1);3*2^（n+1）+2/3an-1&#47，an=2^n*bn=2^2n－2^n(2) 把an=2^2n－2^n代入Sn=4&#47
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