1)首先證明
(4^n+1)(4^(n+1)-1)>4^n(4^(n+1)+1)
證明：
左－右
=[4^(2n+1)+3*4^n-1] - [4^(2n+1)+4^n]
=2*4^n-1>0
2)
f(n)=(4^n-1)/4^n;
P＝f(1)f(2)f(3)...f(n)
=(4-1)/4*(4^2-1)/4^2 *(4^3-1)/4^3...(4^n-1)/4^n
=(4-1)*(4^2-1)*(4^3-1)*...*(4^n-1)/[4*4^2*4^3*...*4^n]
對(duì)于前兩項(xiàng)3*15=45>34=2*17=2*(4^2+1)
所以
P>2*(4^2+1)*(4^3-1)*...*(4^n-1)/[4*4^2*4^3*...*4^n]
再反復(fù)應(yīng)用1)的結(jié)論
P>2*4^2*(4^3+1)*(4^4-1)*...*(4^n-1)/[4*4^2*4^3*...*4^n]
>2*4^2*4^3*(4^4+1)(4^5-1)*...*(4^n-1)/[4*4^2*4^3*...*4^n]
...
>2*4^2*4^3*4^4*...*4^(n-1)*(4^n+1)/[4*4^2*4^3*...*4^n]
=2*(4^n+1)/(4*4^n)
>2*4^n/(4*4^n)
>1/2