Given the following functions (1) f(n a b c ) = ac if n=1 (2) f(n a b c) = f( n-1 a c b) + f( 1 a b c) + f( n-1 b a c ) if n > 1 Then what is the value f( 2 a b c ) = ?

• f(n a b c ) = ac for n=1
  f(n a b c) = f( n-1 a c b) + f( 1 a b c) + f( n-1 b a c ) for n > 1
  so f( 2 a b c )=f( 2-1 a c b) + f( 1 a b c) + f( 2-1 b a c ) as n=2>1
  =f( 1 a c b) + f( 1 a b c) + f( 1 b a c )
  (put the values of three term from 1st eqn for n=1)
  =ab+ac+bc
• 11 years agoHelpfull: Yes(80) No(0)
• f( 2 a b c)=ab + bc +ac
• 11 years agoHelpfull: Yes(5) No(0)
• the ans is ab+ac+bc
• 11 years agoHelpfull: Yes(2) No(0)
• ans:
  ab+bc+ac
• 11 years agoHelpfull: Yes(2) No(1)
• f(n abc)=f(n-1 acb)+f(1 abc)+f(n-1 bac)
  asked is f(2abc) so put n=2 we get f(2abc)=f(1 acb)+ac+f(bac)
  so by the funct definition middle term is omitted so f(2abc)=ab+ac+bc
• 11 years agoHelpfull: Yes(2) No(0)
• f(2 a b c)=ab+ac+bc
  f(1 a c b)=ab
  f(1 b a c)=bc
• 11 years agoHelpfull: Yes(1) No(2)
• for,f(2abc) = f(2-1abc)+f(1abc)+f(2-1abc)
  =f(1abc)+f(1abc)+f(1abc)
  =3f(1abc)
  =3ac (since, f(nabc)=ac; n=1)
  
• 11 years agoHelpfull: Yes(0) No(7)