Question

在数列{an}中，a2=1/4,且(n-an)an+1=(n-1)an(n∈N*). (1)求a1

在数列{an}中，a2=1/4,且(n-an)an+1=(n-1)an(n∈N*). (1)求a1,a3,a4； (2)猜想an的表达式，并用数学归纳法证明

Answer 1

望采纳 谢谢

Answer 2

(n-an)a(n+1)=(n-1)an
a(n+1) = (n-1)an/ (n- an)
1/a(n+1) = (n-an)/[(n-1)an]
= n/[(n-1)an] - 1/(n-1)
1/[na(n+1)] - 1/[(n-1)an] = -1/n(n-1)
= [ 1/n -1/(n-1) ]
1/[(n-1)an] - 1/[(n-2)a(n-1)] = [ 1/(n-1) - 1/(n-2) ]
1/[(n-1)an] - 1/a2 = 1/(n-1) - 1
1/[(n-1)an] = 1/(n-1) +3
= (3n-2)/(n-1)
an = 1/(3n-2)
ie
an = 1/(3n-2)
By MI
an =1/(3n-2)
(1-a1)a2=0
a1=1
p(1) is true
Assume p(k) is true
ie
ak = 1/(3k-2)
for n=k+1
(k-ak)a(k+1)=(k-1)ak
a(k+1) =(k-1)ak/(k-ak)
=[(k-1)/(3k-2)]/[k-1/(3k-2)]
= (k-1)/(3k^2-2k-1)]
= (k-1)/(3k+1)(k-1)]
=1/(3k+1)
p(k+1) is true
By principle of MI, it is true for all +ve integer n
a1=1
a3=1/(9-2)=1/7
a4=1/(12-2)=1/10

追问

谢谢