+26393 Find \(f(k)\) such that \(\sum \limits_{k=1}^n f(k) = n^3\). :
answer see: https://web2.0calc.com/questions/math-question-please-help_11#r2
We can solve this problem using the Mobius Inversion Formula. (See https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula)
By the Mobius Inversion Formula,
f(k) = k^3 - (k - 1)^3 + (k - 2)^3 - (k - 3)^3 + ...
= 3(k^2 + (k - 1)^2 + (k - 2)^2 + ...) - 6*(k + (k - 1) + (k - 2) + ...) + 3
= 3*k(k + 1)(2k + 1)/6 - 6k(k + 1)/2 + 3
= k^3 - 3/2*k^2 - 5/2*k + 3.
+118687 Hi Guest,
It was NewMember, the asker, who voted your answer down.
(I'll give you my up point but it will only take your answer back to zero, sorry.)
It is the way some ignorant people respond when they do not understand an answer.
They should of course just ask you for further explanation or simply say they do not understand.
To be honest, I do not understand you answer either. Maybe it is right, I do not know.
I am not really asking you to explain I am just stateing a fact.
+26393 Find \(f(k)\) such that \(\sum \limits_{k=1}^n f(k) = n^3\). :
answer see: https://web2.0calc.com/questions/math-question-please-help_11#r2
+114 The reason I disliked your answer is because it was incorrect. It is easy to see that the formula did not work when I plugged in values.
+118687 ok, next time give your reason right from the beginining.
Don't vote people down. Thank them for their interest in you question, explain why you think they are incorrect.
Then maybe you will learn together when someone else answers.
If you are both members you could even become friends.
+118687 To NewMember
I'd like if if you 'had it' and I a appreciate you use of 'thanks' here ........ but so far I see evidence to the contrary.
You have voted answer 1 down AGAIN
You have voted yourself up saying 'got it'.
You have NOT thanked Heureka here for linking you to his answer.
AND
You have NOT thanked Heureka for his answer on the original thread.
Seems you know how to abuse people but you don't know much about common courtesy.
