作业帮an=1/n(n+2),Tn为an数列前n项的和,证明T
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an=1/n(n+2),Tn为an数列前n项的和,证明T
an=1/n(n+2),Tn为an数列前n项的和,证明T
an=1/n(n+2),Tn为an数列前n项的和,证明T
题错了,应为小于3／4；
前三项的和为63／120（1／3＋1／8＋1／15）,已大于1／2.
Tn＝1／（1*3）＋1／（2*4）＋1／（3*5）⋯⋯＋1／（n（n＋2））
＝（1／2）（1／1－1／3）＋（1／2）（1／2－1／4）＋（1／2）（1／3－1／5）＋⋯⋯＋
（1／2）（1／n－1／（n＋2））
＝（1／2）（1／1－1／3＋1／2－1／4＋1／3－1／5＋⋯⋯＋1／n－1／（n＋2））
＝（1／2）（3／2－1／（n－2））
小于3／4
an=1/n(n+2)=1/2 (1/n-1/(n+2))
Tn=a1+a2+...an=1/2(1-1/(1+2)+1/2-1/（2+2）+1/3-1/(3+2)......-1/(n+2))
=1/2(1+1/2-1/(n+1)-1/n+2)=3/4-(1/2)(1/n+1+1/(n+2))
1/(n+1)+1/（n+2）>0,
Tn=3/4-(1/2)1/(n+1)+1/（n+2）<3/4
a(n) = (n+1)/[n(n+1)(n+2)] = [(n+2)-1]/[n(n+1)(n+2)]
= 1/[n(n+1)] - 1/[n(n+1)(n+2)]
= 1/n - 1/(n+1) - (1/2){1/[n(n+1)] - 1/[(n+1)(n+2)] },
t(n) = a(1)+a(2)+a(3)+...+a(n-1)+a(n)
=1/1-1/...
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a(n) = (n+1)/[n(n+1)(n+2)] = [(n+2)-1]/[n(n+1)(n+2)]
= 1/[n(n+1)] - 1/[n(n+1)(n+2)]
= 1/n - 1/(n+1) - (1/2){1/[n(n+1)] - 1/[(n+1)(n+2)] },
t(n) = a(1)+a(2)+a(3)+...+a(n-1)+a(n)
=1/1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/(n-1)-1/n + 1/n-1/(n+1) - (1/2){1/[1*2]-1/[2*3] + 1/[2*3]-1/[3*4] + 1/[3*4]-1/[4*5] + ... + 1/[(n-1)n]-1/[n(n+1)] + 1/[n(n+1)]-1/[(n+1)(n+2)] }
=1/1 - 1/(n+1) - (1/2){1/[1*2] - 1/[(n+1)(n+2)]}
= 1 - 1/(n+1) - 1/4 + (1/2)/[(n+1)(n+2)]
= 3/4 - [2(n+2) - 1]/[2(n+1)(n+2)]
= 3/4 - (2n+3)/[2(n+1)(n+2)]
= 3/4 - (1/2)[1/(n+1) + 1/(n+2)],
{t(n)}单调递增。
1/3 = t(1) < t(n) < 3/4.
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