I am actually bored, lets try solving this one:

Here is a function $f$ such that $f(2) = 60 $ and $ −f(1) + f(2) − f(3) +
f(4)−f(5) \cdots +f(n) = nf(n)$, for all positive integers $ n 1 $ Can you find $ f(11) $ ?

Question

I am actually bored, lets try solving this one:

Here is a function $f$ such that $f(2) = 60 $ and $ −f(1) + f(2) − f(3) +
f(4)−f(5) \cdots +f(n) = nf(n)$, for all positive integers $ n > 1 $ Can you find $ f(11) $ ?

anonymous · Answer

Is this too hard ?

mr.math · Answer

No. Give us some time.

anonymous · Answer

Sure :)

mr.math · Answer

$$f(2)-f(11)=11f(11) \implies 60=12f(11) \implies f(11)=5.$$

anonymous · Answer

Not really.

anonymous · Answer

Good question.  This one takes a while due to its recursive nature ...

anonymous · Answer

Thanks, one could try solving it by general analysis and breaking it into smaller cases.

anonymous · Answer

ksaimouli, please don't post your questions in other questions. Yeah I give up FoolForMath :-( Do you have an answer?

anonymous · Answer

Yes, 10.

anonymous · Answer

Haha what a stupid question. "Does FFM have an answer to a mathematical problem?" is the constant value of "YES" :-P

anonymous · Answer

lol, I am not that smart buddy, check out my username -> FOOOOOL :D

anonymous · Answer

I don't believe THAT for a single second! ;)

anonymous · Answer

Yes, believe it after that single second past :P

anonymous · Answer

LIES!

mr.math · Answer

Wait, where did I go wrong. We can easily show that f(1)=0, and that f(2n-1)=f(2n) for $n\ge 2$. From here we can reduce $-f(1)+f(2)-f(3)+..-f(11)=f(2)-f(11)$. Doesn't this sound right?

anonymous · Answer

This is a GRE problem, and $ 5 $ is not even in the four given option. Something wrong in you analysis I believe, I got different f(2k) and f(2k+1) for n=2k and n=2k+1 respectively.

mr.math · Answer

Could you give a counterexample?

mr.math · Answer

Oh the definition says for all n>1.

mr.math · Answer

Isn't f(1) given by the definition?

nikvist · Answer

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