(1)在数列an中 a1=1, a(n+1)=10(an)^2,求an 在数列an中，a1=1,an+1=(1+1/n)an+n+1...

(1)
a(n+1)=10(an)²,两边取lg得：
lg(an+1)=1+2lgan
令bn=lgan
b(n+1)=2bn+1
b(n+1)+1=2[(bn)+1]
{bn +1}是等比数列；
首项是：b1 +1=1,q=2
bn +1=2^(n-1)
bn=2^(n-1)-1
an=10^[2^(n-1)-1]
(2)
a(n+1)=3an+(2n-1)*3^n
两边同除以：3^(n+1)
a(n+1)/3^(n+1)=an/3^n+(2n-1)/3
令bn=an/3^n
b(n+1)=bn+(2n-1)/3
b1=(1/3)
b2-b1=(1/3)*1
b3-b2=(1/3)*3
.......................
bn-b(n-1)=(1/3)*(2n-3)
把这n个式子相加得：
bn=(1/3)+(1/3)[1+3+5+...+(2n-3)=(1/3)+n²/3
an/3^n=(1/3)+n²/3
an=3^(n-1)+n²*3^(n-1)

(1) 设 bn = 10 an
则 b(n+1) = (bn)^2
故 bn = b1^(2^(n-1)) = 10^(2^(n-1))
an = 10^(2^(n-1) - 1)

(2) 设 an = bn * 3^(n-1) 则
b(n+1) - b(n) = 2n - 1
b(n+1) - b(1) = (2n-1) + (2(n-1)+1) + ... + (2 - 1) = n^2
所以
bn = (n-1)^2+1
an = ((n-1)^2 +1) * 3^(n-1)

a(n+1)=2an+2^n
同除以2^n
a(n+1)/2^n=2an/2^n+1
a(n+1)/2^n-an/2^(n-1)=1
所以数列｛an/2^(n-1)｝为以1为公差的等差数列
a1/2^0=1
an/2^(n-1)=1+(n-1)*1=n
所以an
=
n2^(n-1)
sn=1*2^0+2*2^1+3*2^2+....+
n2^(n-1)
2sn=
1*2^1+2*2^2+....+(n-1)2^(n-1)+n2^n
用2式－1式
sn=-1-2^1-2^2-....2^(n-1)+n2^n
=-1-(2+2^2+2^3+...+2^(n-1))+n2^n
=(n-1)2^n+1
(错位相消法)

在数列an中，a1=1,a(n+1)=an/(an+1)~

(1)
a(n+1)=an/(an+1)

1/a(n+1) = (an+1)/an
1/a(n+1) -1/an = 1
=>(1/an)是等差数列

1/an -1/a1= n-1
1/an =n
an =1/n
(2)
bn =1/(2^n.an)
= (1/2)[n(1/2)^(n-1)]
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=1/2
summation(i:1->n)i.(1/2)^(i-1)
=4[n.(1/2)^(n+1) - (n+1).(1/2)^n + 1]
=4(1- (n+2). (1/2)^(n+1) )
Sn = b1+b2+...+bn
= (1/2). {summation(i:1->n)i.(1/2)^(i-1)}
= 2(1- (n+2). (1/2)^(n+1) )

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+1)=b(n) + 1/2^n, b(1)=a(1)/1=1.
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=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)/n,
a(n)=n[2^n-1]/2^(n-1)=2n - n/2^(n-1).

s(n)=n(n+1)-[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)=2/1 + 2/1 + 3/2 + ... + (n-1)/2^(n-3) + n/2^(n-2),
t(n)=2t(n)-t(n)=2/1 + 1/1 + 1/2 + ... + 1/2^(n-2) - n/2^(n-1)=2+[1-1/2^(n-1)]/[1-1/2] - n/2^(n-1)
=2 + 2[1-1/2^(n-1)] - n/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)