Question

a problems #4
my entrance exam pb:

Answer 1

equation?

Answer 2

when\[ f(x+y) = f(x)+f(y)+4xy\] and \[f(1)=4\] find \[f(20)\]
P.S. as ever, if you cannot do it within 10 minutes, give me your medal;))

Answer 3

be calm man

Answer 4

fx+fy = fx +fy +4xy i really dont know this lol

Answer 5

i go learn it

Answer 6

lol

Answer 7

thought this would take you guy some time :P

Answer 8

you know i remember the hard time i'll have to write down from f(1)to f(20) :((

hope you guy have a better solution

Answer 9

Apparently f(2) = 12, f(4) = 40, f(8) = 144, f(16) = 83,232, and so finally f(20) = f(16 + 4) = 13,400,392

Answer 10

huh? i dont understand.
could you explain it more clearly? jemurray
btw as far as i can recall i got just 3 digit number

Answer 11

Oh, oops, I made a typo in my calculator :) Just a second...

Answer 12

nononooooooooooooooooooooooooooooooooooooooooooo im hungry:(

Answer 13

f(16) = 416, f(20) = f(4 + 16) = 712

Answer 14

I get 840
\[f(n)=2n(n+1)\]

Answer 15

Here's one way to do it

f(x+y)=f(x)+f(y)+4xy

f(20)= f(1+19)= f(1)+4*1*19 + f(19) = 4+4*19 + f(19)
f(20)= 4*20 + f(19)
continue the pattern
f(20)= 4*(1+2+...+20)= 4*20*21/2= 840

Answer 16

Here is how I did it, I computed \( f(1+1)=f(2)\) , then \( f(3)=f(2)+f(1) \) then \(f(5)\),\( f(10) \) and finally \( f(20)\).