a2(n+1)-a2n=a(2n+2)-an=2d
故{a2n}的公差是2d
{a[2(n+1)-1]-a2(n+1)}-[a(2n-1)+a2n]=a(2n+1)-a(2n+2)-a(2n-1)+a2n=0
{a2n-1+a2n}的公差为0
这是简单的子数列问题a2(n+1)-a2n=a(2n+2)-an=2d??这步写错了吧?是a2(n+1)-a2n=a(2n+2)-a2n=2d吧...{a[2(n+1)-1]-a2(n+1)}-[a(2n-1)+a2n]=a(2n+1)-a(2n+2)-a(2n-1)+a2n=0{a2n-1+a2n}的公差为0....答案是4d啊。。。第一个应当没错a2(n+1)-a2n=a(2n+2)-an=a(2n+1)+d-a2n=2d第二个我看错符号了，嘻嘻，太急了，不好意思{a[2(n+1)-1]+a2(n+1)}-[a(2n-1)+a2n]=a(2n+1)+a(2n+2)-a(2n-1)-a2n=2d+2d=4d