f(f(n))+f(n)=2n+3, f(0)=1,
find f(2012)?

• n=0
  f(f(0)+f(0)=2*0+3
  f(1)+1=3
  f(1)=2
  n=1
  f(f(1))+f(1)=2*1+3
  f(2)+2=5
  f(2)=3
  n=2
  f(f(2))+f(2)=2*2+3
  f(3)+3=7
  f(3)=4
  similarly
  f(2012)=2013
• 12 years agoHelpfull: Yes(37) No(3)
• f (f(0)) + f(0) = 2(0) + 3, f(1) = 3-1 = 2, f(1) = 2
  f(f(1)) + f(1) = 2(1) + 3, f(2) = 5-2 = 3, f(2) = 3
  f(f(2)) + f(2) = 2(2) + 3, f(3) = 7-3 = 4, f(3) = 4
  That is, f(n)=n+1
  So, f(2012) = 2013
• 7 years agoHelpfull: Yes(2) No(0)
• f[f(n))+f(n)=2n+3
  as they mention f[0]=1,
  apply in the question-> f(f(0))+f(0)=2(0)+3
  -> f(1)+1=3 (sincef[0]=1)
  therfore f[1]=2
  similarly f[1] ,
  f[f[1]]+f[1]=2(1)+3
  f(2)+2=2(1)+3
  f(2)=5-2
  f(2)=3
  like that for every f, one value is increased
  f[2012]=2013
• 7 years agoHelpfull: Yes(1) No(0)