4的2n－1次方與16的n次方的和
=4^(2n-1)+16^n
=(2)^[2*(2n-1)]+2^(4*n)
=2^(4n-2)+2^(4n)
=[2^(4n-2)]*(1+2^2)
=5*2^(4n-2)
2的4n＋1次方 減去4的2n－1次方與16的n次方的和
=2^(4n+1)-5*2^(4n-2)
=8*2^(4n-2)-5*2^(4n-2)
=3*2^(4n-2)