∵an·a(n+1) + 2a(n+1) + 1 = 0
∴an·a(n+1) + a(n+1) + 1 = -a(n+1)
an·a(n+1) + an + a(n+1) + 1 = an - a(n+1)
(an + 1)·(a(n+1) + 1)= an - a(n+1)
即:
an - a(n+1)
———————————— = 1
(an + 1)·(a(n+1) + 1)
则:
1 1
—————— — ——————
a(n+1) + 1 an + 1
an + 1 - 【a(n+1) - 1】
= ————————————
(an + 1)·(a(n+1) + 1)
an - a(n+1)
= ————————————
(an + 1)·(a(n+1) + 1)
= 1
∵a1=1
∴1/(an + 1)=1/(1+1)=1/2
则数列{1/(an + 1)}为首项是1/2,公差是1的等差数列.
☆☆简单题干,求证等差数列的,急!
☆☆简单题干,求证等差数列的,急!
已知数列{an}满足 a1=1,an*a(n+1)+ 2a(n+1) + 1 = 0 (n∈N+),
求数列{1/(an+1)}为等差数列
已知数列{an}满足 a1=1,an*a(n+1)+ 2a(n+1) + 1 = 0 (n∈N+),
求数列{1/(an+1)}为等差数列
数学人气:758 ℃时间:2020-10-01 19:02:49
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