已知数列{an}中满足(An+1-An)(An+1+An)=16,且a1=1,an
问题描述:
已知数列{an}中满足(An+1-An)(An+1+An)=16,且a1=1,an
答
1.[a(n+1)-an][a(n+1)+an]=16a(n+1)²-an²=16,为定值.a1=1 a1²=1数列{an²}是以1为首项,16为公差的等差数列.2.an²=1+16(n-1)=16n-15n≥2时,16n-15≥16×2-15=17>0an