数列【An】的前n项和Sn=1/2n的平方—2n[n属于N*],数列【Bn】满足Bn=An+1/An[n属于N*],判断数列,【An】是

an=Sn-Sn-1 = (1/2) n^2-2n-1/2(n-1)^2+2(n-1) = 1/2(2n-1)12=n-5/2 (n≥2)
a1=s1 = -3/2 综上,an = n-5/2 [n属于N*],an-an-1 = 1是常数
这是以-3/2为首项,1为公差的等差数列
应该是 bn =a(n+1)/an 吧,数列题还是要写清楚点- -
bn = (n-5/2 + 1)/(n-5/2) = 1+1/(n-5/2)
此时,n=3时,bn最大值 = 3
n=2 时 bn最小值 = -1