(1)设数列{an}满足{a1=-1; { an=a(n-1)^2+1(n>1) 写出它的前四项

2024-11-22 19:09:11
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回答(1):

(1)
an=(a(n-1))^2+1
a1=-1
n=2
a2= (a1)^2+1 = 1+1=2
n=3
a3=(a2)^2+1=4+1=5
n=4
a4 = (a3)^2=1= 25+1=26
(2)

7,4,3,..., (n+6)/n,...
n=10
a10= (10+6)/10= 16/10=8/5

53/50= (n+6)/n
53n=50n+300
n=100

这个数列有多少个整数项=3个
a1=7
a2=4
a3=5

(n+6)/n = n/3
3(n+6) = n^2
n^2-3n-18=0
(n-6)(n+3)=0
n=6