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}的公差為0
這是簡(jiǎn)單的子數(shù)列問題a2(n+1)-a2n=a(2n+2)-an=2d??這步寫錯(cuò)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}的公差為0....答案是4d啊。。。第一個(gè)應(yīng)當(dāng)沒錯(cuò)a2(n+1)-a2n=a(2n+2)-an=a(2n+1)+d-a2n=2d第二個(gè)我看錯(cuò)符號(hào)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