在数列{an}中,已知a1=1.a2=1/2, 2/an=1/a(n-1) + 1/a(n+1)？

Question

如题所述

Answer 1

解,
1）求an
设 bn = 1/an
则 2bn = b(n-1) + b(n+1),即 bn 为等差数列.
b1 = 1/a1 = 1,b2 = 1/a2 = 2
显然 bn = n,an = 1/n
a3,a4就不写了吧
2)新数列 ana(n+1)
设此数列为 = ana(n+1) = 1/n(n+1) = 1/n - 1/(n+1)
则 c1 = 1/2
c2 = 1/2 *1/3 = 1/2
sn = 1/2 + (1/2- 1/3) + (1/3-1/4) + ...+ (1/n - 1/(n+1))
当n =1时,sn = 1/2
当n>1时,sn = 1 - 1/(n+1)
所以 1/2≤Sn,2,在数列{an}中,已知a1=1.a2=1/2, 2/an=1/a(n-1) + 1/a(n+1)
求a3,a4,求数列an的通项公式, 若数列{ana(n+1)}的前n项和为Sn.求证：1/2≤Sn