Question

数列{an}满足a1=1.an=an+1(1+2an)(n∈N)

数列{an}满足a1=1.an=an+1(1+2an)(n∈N)(1){1/an}是等差数列,(2)若a1a2+a2a3+……＋anan+1>16/33求n的取值范围内

Answer 1

[[[1]]]
由题设a1=1,且an=[a(n+1)]×(1+2an)
可得: a1=1, a2=1/3,
且1/[a(n+1)]=[1/(an)]+2 1/(a2)=(1/a1)+2
∴该数列是首项为1,公差为2的等差数列
∴通项1/an=1+2(n-1)=2n-1
∴an=1/(2n-1). n=1,2,3,,,,,
[[[2]]]]
由上面可知,
2(an)×[a(n+1)]=[1/(2n-1)]×[1/(2n+1)]=[1/(2n-1)]-[1/(2n+1)].
分别取n=1,2,3,4,,,,,n
把所得式子累加,可得
2{a1a2+a2a3+...+ana(n+1)]=1-[1/(2n+1)]=(2n)/(2n+1)
∴a1a2+a2a3+...+ana(n+1)=n/(2n+1)＞16/33
∴n＞16
∴n的最小值为17

Answer 2

1.a(n)=a(n+1)(1+2a(n))
1/a(n+1)=1/an*(1+2an)=2+1/an
an=1/(2n-1)
2.a(n+1)a（n）=(1/(2n-1))*(1/(2n+1))
a1a2+a2a3+……＋anan+1=0.5（1-1/(2n+1))>16/33
n>16