f(f(n))=2n+3 and f(0)=1, what is the value of f(2012)?

• the question is,
  f(f(n))+f(n)=2n+3 ,f(0)=1 ,what is the value of f(2012)?
  ans = 2013
  because from given equation getting
  f(0)=1,f(1)=2,f(2)=3............
• 11 years agoHelpfull: Yes(23) No(4)
• Put n = 0
  Then f(f(0))+f(0) = 2(0) + 3 ⇒ f(1) + 1 = 3 ⇒ f(1) = 2
  Put n = 1
  f(f(1)) + f(1) = 2(1) + 3 ⇒ f(2) + 2 = 5 ⇒f(2) = 3
  Put n = 2
  f(f(2)) + f(2) = 2(2) + 3 ⇒ f(3) + 3 = 7 ⇒ f(3) = 4
  ......
  f(2012) = 2013
• 11 years agoHelpfull: Yes(8) No(0)
• Given f(f(n))=2n+3
  Let f(n)=m
  Then,f(m)=2m+3
  Now,f(2012)=2*2012+3=4027
  So,m=4027
  Now,f(m)=f(4027)=2*4027+3=8057
  Hence,8057 is the answer.
• 11 years agoHelpfull: Yes(5) No(17)
• 2013 will be answer..
  .
• 11 years agoHelpfull: Yes(3) No(1)
• f(2014)=2*2012+3=4027
  and f(4027)=2*4027+3=8057 So answer will be 8057.
• 11 years agoHelpfull: Yes(0) No(3)
• 2013
• 11 years agoHelpfull: Yes(0) No(2)
• Put n = 0
  Then f(f(0))+f(0) = 2(0) + 3 ⇒ f(1) + 1 = 3 ⇒ f(1) = 2
  Put n = 1
  f(f(1)) + f(1) = 2(1) + 3 ⇒ f(2) + 2 = 5 ⇒f(2) = 3
  Put n = 2
  f(f(2)) + f(2) = 2(2) + 3 ⇒ f(3) + 3 = 7 ⇒ f(3) = 4
  ......
  f(2012) = 2013
• 11 years agoHelpfull: Yes(0) No(0)