a(n+1)+an=4n-3
an-a(n-1)=4(n-1)-3
两式相减,得到
a(n+1)-a(n-1)=4
也就是说a1,a3,a5,...a(2k-1)成等差数列,
a2,a4,a6,...a(2k)成等差数列,公差都为4,
a(2k-1)=a1+(k-1)d=2+4(k-1)=4k-2
当n=1,a2+a1=4-3=1,a2=-1
a(2k)=a2+(k-1)d=-1+4(k-1)=4k-5
an的通项公式为
an=4k-2 (n=2k-1)
4k-5 (n=2k)