If f(x +y)= f(x) + f(y) + 2xy. find f(8).

Question

  x,y∈N and f(1)=4

Answer 1

Jon P. has given a simple solution that works for f(8) and other powers of 2.  So, if you want a quick solution, read his answer.  Shradha S. has given a more general solution [that could answer, for example, the question, "What is f(13)?"].

Here is the recursive definition of the function (by letting y=1): 

     f(x+1) = f(x) + f(1) + 2x(1)

The function is:

    f(x+1) = f(x) + 4 + 2.x       when x>0

    f(1) = 4                              when x=0

 

This expands (note: this is easy for a computer) to:

  f(8)=f(7+1)=    f(7)                          +    f(1)+2*7*1          |    4 + 14

    f(7)=f(6+1) = f(6)                         +     f(1)+2*6*1          |    4 + 12

      f(6)=f(5+1) = f(5)                       +     f(1)+2*5*1          |    4 + 10

         f(5)=f(4+1) = f(4)                    +     f(1)+2*4*1          |    4 +   8

           f(4)=f(3+1) = f(3)                  +     f(1)+2*3*1          |    4 +   6

             f(3)=f(2+1) = f(2)                +     f(1)+2*2*1          |    4 +   4

              f(2)=f(1+1) = f(1)               +     f(1)+2*1*1          |    4 +   2

                f(1) =   4                                                             |    4

                                                                                            ------------------

                                                                                              32 +  56 =  88

Answer 2

Start by calculating f(2):

 

f(2) = f(1 + 1) = f(1) + f(1) + 2*1*1 = 4 + 4 + 2 = 10

 

Now f(4)...

 

f(4) = f(2 + 2) = f(2) + f(2) + 2*2*2 = 10 + 10 + 8 = 28

 

Now f(8)...

 

f(8) = f(4 + 4) = f(4) + f(4) + 2*4*4 = 28 + 28 + 32 = 88

 

 

Answer 3

Given  f(x +y)= f(x) + f(y) + 2xy    and    f(1)=4

 

As we know the value of f(1) only we break f(8) as f(7+1) and expand using the definition given, then we get the term f(7). again we break f(7) as f(6+1) and expand and continuing in the same way finally we wil get terms contaning f(1) +some numbers

 

f(8) = f(7+1) = f(7) + f(1) + 2*7*1

                    =f(6+1) +f(1) +2*7

                    =f(6) + f(1) + 2*6*1+f(1) +2*7

                    =f(6)+2*f(1) + 2*6+2*7

                    =f(6)+2*f(1) +2*6+2*7

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proceeding in similar manner we finally get

                    =8*f(1) + 2*1+2*2+2*3+2*4+2*5+2*6+2*7

                    =8*4 + 2+4+6+8+10+12+14                                ;As f(1)=4

                    =32+ 56

                    =88