1,a(1)=2>0,a(n+1)=3[a(n)]^2>0. a(n)>0.
ln[a(n+1)]=ln[a(n)]^2 + ln(3) = 2ln[a(n)] + ln(3)
ln[a(n+1)] + ln(3) = 2{ln[a(n)] + ln(3) }
{ln[a(n)] + ln(3)}是首项为ln[a(1)] + ln(3)=ln(6), 公比为2的等比数列.
ln[a(n)]+ln(3)=ln(6)*2^(n-1)=ln{6^[2^(n-1)]} = ln[3a(n)],,
6^[2^(n-1)] = 3a(n)
a(n)=(1/3)*6^[2^(n-1)].
2,a(1)=2,
a(2)=3/(-1)=-3,
a(3)=(-2)/(4)=-1/2
a(4)=(1/2)/(3/2)=1/3
a(5)=(4/3)/(2/3)=2
...
a(4n-3)=2,
a(4n-2)=-3
a(4n-1)=-1/2
a(4n)=1/3
3,a(n+1)=a(n)+2n=a(n)+n(n+1)-(n-1)n,
a(n+1)-n(n+1)=a(n)-(n-1)n=...=a(1)-0=33,
a(n)=33+(n-1)n
a(n)/n = 33/n + n - 1 >= -1 + 2*(33/n*n)^(1/2) = -1 + 2(33)^(1/2),
等号成立当且仅当33/n=n, n^2=33.
a(5)/5=33/5 + 4 = 53/5 = 10+3/5
a(6)/6=33/6 + 5 = 63/6 = 10 + 3/6<a(5)/5
因此,a(n)/n的最小值=a(5)/5=53/5
4,若a(5)为奇数,则1=a(6)=3a(5)+1,a(5)=0矛盾.
因此,a(5)为偶数.1=a(6)=a(5)/2, a(5)=2.
若a(4)为奇数,则2=a(5)=3a(4)+1, 1=3a(4), a(4)不为整数,矛盾.
因此,a(4)为偶数.2=a(5)=a(4)/2, a(4)=4.
(4.1)若a(3)为奇数,则4=a(4)=3a(3)+1, a(3)=1.
若a(2)为奇数,则1=a(3)=3a(2)+1,a(2)=0矛盾.
因此,a(2)为偶数.1=a(3)=a(2)/2, a(2)=2.
若a(1)为奇数,则2=a(2)=3a(1)+1, 1=3a(1), a(1)不为整数,矛盾.
因此,a(1)为偶数.2=a(2)=a(1)/2, a(1)=4=m.
(4.2)若a(3)为偶数,则4=a(4)=a(3)/2, a(3)=8.
若a(2)为奇数,则8=a(3)=3a(2)+1,a(2)不为整数,矛盾.
因此,a(2)为偶数.8=a(3)=a(2)/2, a(2)=16.
若a(1)为奇数,则16=a(2)=3a(1)+1, a(1)=5=m.
若a(1)为偶数.则16=a(2)=a(1)/2, a(1)=32=m.
综合,有,m所有可能取值为4,5,32.
5, na(n+1)=s(n)+n(n+1)=n[s(n+1)-s(n)],
ns(n+1)=(n+1)s(n)+n(n+1),
s(n+1)/(n+1)=s(n)/n + 1,
{s(n)/n}是首项为s(1)/1=a(1)=1,公差为1的等差数列.
s(n)/n=1+(n-1)=n,
s(n)=n^2
na(n+1)=s(n)+n(n+1),
a(n+1)=n+n+1=2n+1=2(n+1)-1,
a(n)=2n-1.
t(n)=(4/5)^n*s(n)=(4/5)^n*n^2>0.
lim_{n->正无穷}t(n)=0
因此,不存在正整数m，对一切正整数n总有Tn小于等于Tm.
6,a(n+1)=(n+1)a(n)/n + (n+1)/2^n
a(n+1)/(n+1)=a(n)/n + 1/2^n,
b(n)=a(n)/n,
b(n+1)=b(n)+1/2^n,
2^nb(n+1)=2*2^(n-1)b(n)+1,
2^nb(n+1)+1 = 2[2^(n-1)b(n)+1]
{2^(n-1)b(n)+1}是首项为b(1)+1=a(1)/1+1=2,公比为2的等比数列.
2^(n-1)b(n)+1=2^n,
b(n)=[2^n-1]/2^(n-1),
a(n)=nb(n)=n[2^n-1]/2^(n-1)=2n - n/2^(n-1)
s(n)=2[1+2+...+n] - [1/1+2/2+3/2^2+...+(n-1)/2^(n-2)+n/2^(n-1)]
=n(n+1) - t(n)
t(n)=1/1+2/2 + 3/2^2 + ... + (n-1)/2^(n-2) + n/2^(n-1)
2t(n)=2 + 2/1 + 3/2 + ... + (n-1)/2^(n-3)+n/2^(n-2),
t(n)=2t(n)-t(n)=2 + 1/1 + 1/2 + ... + 1/2^(n-2) - n/2^(n-1)
=2-n/2^(n-1) + 2[1-1/2^(n-1)]
=4 -(n+2)/2^(n-1)
s(n)=n(n+1)-t(n)=n(n+1)-4+(n+2)/2^(n-1)