Fool's problem of the day,

If $f(x,y+1) = f(x,y)+x+1$ and $f(y,0) =y \forall x,y \in \mathbb{Z} $. Then can you find $f(7,6)$?

Question

Fool's problem of the day,


If $f(x,y+1) = f(x,y)+x+1$ and $f(y,0) =y \forall x,y \in \mathbb{Z} $. Then can you find $f(7,6)$?

anonymous · Answer

PS: This is a very easy problem, I am uploading on request of @Callisto.

anonymous · Answer

Well, I gave you a headline for it ;)

callisto · Answer

Nah.. Thanks.. :P

anonymous · Answer

Haha, Thanks CS :D , now I wish to have posted something harder ;)

experimentx · Answer

f(7,0) = 7
f(7,1) = 7+8 = 16
f(7,2) = 16 + 8

..

ap??

experimentx · Answer

f(7,6) = 7+6*8 = 55??

anonymous · Answer

Yes,

$$ f(7, 6) = f(7, 5) + 7 + 1 = f(7, 5) + 8 $$
$$= f(7, 4) + 16 = f(7, 3) + 24$$
$$ = f(7, 2) + 32 = f(7, 1) + 40 $$
$$ = f(7, 0) + 48 = 7 + 48 = 55 $$

anonymous · Answer

f(x,y+1)=f(x,y)+x+1 
f(7,6) = f(7,5) +7 +1
         =f(7,4)+7+1+7+1 = f(7,4)+2(7)+2(1)
         = f(7,3)+7+1+7+1+7+1 = f(7,3)+3(7)+3(1)
              :
              :
          = f(7,0) +6(7)+6(1)
           = 7+42+6
           =55

anonymous · Answer

I think I will post something harder now ;)

anonymous · Answer

Okay somewhat interesting : http://openstudy.com/study#/updates/4f746e7ae4b0b478589db2d5