s(n+ 1)+4S(n- 1)= 5Sn；

S(n+ 1)-Sn = 4[Sn-S(n- 1)]；

a(n+ 1)= 4an；

So {an} is a geometric series with 2 as the first term and a common ratio of 4.

an=a 1*4^(n- 1)=2*2^(2n-2)=2^(2n- 1)；

2.

Let pn = log2 (an) = 2n-1;

Tn=P 1+P2+...+Pn= 1+3+5+...+2n- 1=n？ (arithmetic progression summation);

3.

( 1- 1/T2)( 1- 1/T3).....( 1- 1/Tn)；

=( 1- 1/2? )( 1- 1/3? ).....( 1- 1/n？ );

=3/4*8/9* 15/ 16*24/25*........；

Pay attention to the law;

3/4*8/9=2/3=4/6。 . . . . . n = 2；

3/4*8/9* 15/ 16=5/8...........n = 3；

3/4*8/9* 15/ 16*24/25=3/5=6/ 10.............n = 4；

3/4*8/9* 15/ 16*24/25*35/36=7/ 12.............n = 5；

∴ Original formula = (n+2)/(2n+2);

When n= 100;

Original formula =102/202 = 51101;

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