(1)充分性.若{an}是等差数列,设公差为d
则 1/an -1/a(n+1)=[a(n+1)-an]/ana(n+1)=d/ana(n+1)
所以1/a1a2+1/a2a3+……+1/ana(n+1)
=[1/a1 -1/a2 +1/a2 -1/a3+...+1/an -1/a(n+1)]/d
=[1/a1 -1/a(n+1)]/d=[a(n+1) -a1]/[a1a(n+1)d]=n/a1a(n+1)
(2)必要性.若 1/a1a2+1/a2a3+……+1/ana(n+1)=n/a1a(n+1),
则 1/a1a2+1/a2a3+……+1/a(n-1)an=(n-1)/a1an
两式相减,得 1/ana(n+1)=n/a1a(n+1) -(n-1)/a1an,
通分,去分母,得 a1=nan -(n-1)a(n+1),
a(n+1) -a1=n[a(n+1)-an]
所以 an -a1=(n-1)[an -a(n-1)]
相减,得 (n-1)[a(n+1)-an]=(n-1)[an -a(n-1)]
a(n+1) -an =an-a(n-1),{an}是等差数列.
以上回答你满意么?