由f(x)=2x-1,g(x)=-2x,点(a2n-1,a2n)在函数y=f(x)或y=g(x)的图象上,知a2n=2a2n-1-1,或a2n=-2a2n-1,当a2n=2a2n-1-1时,a2n-1=2(a2n-1-1),由数列{an} (n∈N*)的各项都是整数,且当n为偶数时,,知,,由此能够求出S80.
【解析】
∵f(x)=2x-1,g(x)=-2x,
点(a2n-1,a2n)在函数y=f(x)或y=g(x)的图象上,
∴a2n=2a2n-1-1,或a2n=-2a2n-1,
∵当n为偶数时,,
∴当a2n=2a2n-1-1时,2a2n-1=a2n+1=n+1,
∴,
令n=2k-1,k∈N*,则a4k-3=
2k-1+1
2
=k,即a1,a5,a9,…,成首项为1,公差为1的等差数列;
当a2n=-2a2n-1时,a2n-1=-
n
2
,
所以n为偶数时,a2n-1=-
n
2
,
令n=2k′,k′∈N*,则a4k′-1=-
2k′
2
=-k′,即a3,a7,a11,…,成首项为-1,公差为-1的等差数列;
所以S4n=S奇+S偶=[(1+2+3+…+n)+(-1-2-3-…-n)]+(1+2+3+4+…+2n)=
2n(1+2n)
2
=2n2+n.
∴S80=2×202+20=820.
故答案为:820.