Question

求大神帮忙，高一数学，数列题，详细过程，在线，急

Answer 1

(1)
a1=3/5
a(n+1)= 3an/(4an+1)
1/(an+1) = 4/3 + (1/3)(1/an)
1/(a(n+1) - 2 = (1/3) ( 1/an - 2)
=> {1/an - 2} 是等比数列 , q= 1/3
(2)
1/an - 2 = (1/3)^(n-1) .( 1/a1 - 2 )
=2 .(1/3)^n
1/an = 2 + 2 .(1/3)^n
Sn =1/a1+1/a2+...+1/an
= 2n + (2/3) ( 1- (1/3)^n ) /(1- 1/3)
= 2n + ( 1- (1/3)^n )
Sn < 100
2n +1 <100
max n =44

Answer 2

（1）、a(n+1)=3an/(4an+1),
取倒数得：1/ a(n+1)=( 4an+1)/(3an)
即有1/ a(n+1)=4/3+1/(3an)
设1/an=bn,上式可化为b(n+1)= 4/3+(1/3)bn
则b(n+1)-2=1/3(bn-2)
所以数列{bn-2}是公比为1/3的等比数列,其首项为b1-2=1/a1-2=-1/3
（2）、
bn-2=-1/3•(1/3)^(n-1)
即1/an-2=-1/3•(1/3)^(n-1)
化简得 1/an=2-(1/3)^n
Sn=1/a1+1/a2+···+1/an
=2-(1/3)^1+2-(1/3)^2+···+2-(1/3)^n
=2n-[1/3-(1/3)^n·1/3]/(1-1/3)
=2n-1/2(1/3)^n+1/2＜100
∴nmax=50

Answer 3