求解两道数学题

第一题:首先把Sn-1式子写出来，然后Sn-Sn-1得到an，化简得an+1=2an,即an是等比数列，其中a1=2,等比是2，那么an=2的n次方，则bn=log2(an)=n,那么1/(bnbn+1)=1/(n*(n+1)),所以前n项和等于
1/(1*2)+1/(2*3)+...+1/(n*(n+1))=1-1/2+1/2-1/3+...+1/n-1/(n+1)=1-1/(n+1)
第二题：
1：a1,a4=a1+3d,a13=a1+12d,S9=9a1+9*8/2d=9a1+36d=99.
因为a1,a4,a13成等比，用等比的性质，即：a4/a1=a13/a4,化简得3d=2a1,带入S9的式子，得到33a1=99,则a1=3,d=2，则an=3+2*(n-1)
2:bn=3+2*(2的n次方-1）+n=2的(n+1)次方+n+1,把这个式子拆成两个式子得到一个等比数列和等差数列，分别求和得到Tn的式子。

(6)
a1=2
Sn = ( 1- 1/2^n).a(n+1)
bn=log<2>an
cn = 1/[bn.b(n+1)]
Tn = c1+c2+...+cn = ?
solution:
for n>=2
an = Sn -S(n-1)
=( 1- 1/2^n).a(n+1) - [ 1- 1/2^(n-1) ].an
2^n.an = ( 2^n- 1).a(n+1) - ( 2^n - 2 ).an
2(2^n -1)an = ( 2^n- 1).a(n+1)
a(n+1) = 2an
an = 2^(n-1) . a1
=2^n
bn=log<2>an =n
cn = 1/[bn.b(n+1)]
=1/[n(n+1)]
=1/n - 1/(n+1)
Tn
= c1+c2+...+cn
=1 - 1/(n+1)
=n/(n+1)
(7)
(I)
an =a1+(n-1)d
Sn=a1+a2+...+an
S9 =99
9(a1+4d)=99
a1+4d=11 (1)
a1.a13= (a4)^2
a1.(a1+12d)= (a1+3d)^2
(11-4d)(11+8d) =(11-d)^2
121+44d-32d^2 =121-22d+d^2
33d^2-66d=0
d(d-2)=0
d=2
from (1)
a1+4d=11
a1+8=11
a1=3
an = 3+2(n-1) = 2n+1
(II)
bn
= a(2^n) +n

= 2(2^n) +1+n
=2^(n+1) + n+1
Tn
=b1+b2+...+bn

=4(2^n -1) + (1/2)n(n+3)