数列an中,a1=1,a2=2且a(n+2)-an=1+(-1)^n求s100

递推公式：
a(n+2)=an+1+(-1)^n
a(n+1)=a（n-1）+1+(-1)^(n-1)
a(n)=a(n-2)+1+(-1)^(n-2)
a(n-1)=a(n-3)+1+(-1)^(n-3)
......
a(5)=a(3)+1+(-1)^3
a(4)=a(2)+1+(-1)^2
a(3)=a(1)+1+(-1)^1
两边相加相消：
a(n+2)+a(n+1)=a(2)+a(1)+n-(1-(-1)^n)(1+1)
=2+1+n-[1-(-1)^n]/2
a(n+1)+a(n)=3+(n-1)-[1-(-1)^(n-1)]/2
s100=
a100+a99
+a98+a97
+...
+a2+a1
=3*99+[0+1+2+..+98]-99/2+1/2*[(-1)^0+(-1)^1+...+(-1)^98]
=3*99+99*98/2+1/2
=99*52+1/2
=5148.5