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Find \(f(k)\) Such That: \(\sum_{k=1}^n f(k) = n^3.\)

 Jan 5, 2020

Best Answer 

 #5
avatar+24344 
+5

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

 

laugh

 Jan 6, 2020
 #1
avatar
+1

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.

 Jan 6, 2020
 #2
avatar+2824 
+1

Why is the answer disliked? What is wrong with it?

CalculatorUser  Jan 6, 2020
 #3
avatar+108624 
+1

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.

Melody  Jan 6, 2020
 #4
avatar+108624 
+1

I will test your answer,

Testing for n=1

 

\(f(k)=k^3-\frac{3k^2}{2}-\frac{5k}{2}+3\\~\\ f(1)=1^3-\frac{3*1^2}{2}-\frac{5*1}{2}+3\\ \quad f(1)=1-\frac{3}{2}-\frac{5}{2}+3\\ \quad f(1)=0\ne 1^3 \)

 

 

Maybe I made a mistake but your formula does not appear to work.

Melody  Jan 6, 2020
 #5
avatar+24344 
+5
Best Answer

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

 

laugh

heureka Jan 6, 2020
 #6
avatar+108624 
+2

Thanks for showing us Heureka.

That is a great solution!

 Jan 6, 2020
 #7
avatar+24344 
+3

Thank you, Melody !

 

laugh

heureka  Jan 6, 2020
 #8
avatar+119 
0

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.

 Jan 6, 2020
 #9
avatar+108624 
+1

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.

Melody  Jan 6, 2020
 #10
avatar+119 
0

Got it! Thanks!

 Jan 7, 2020
 #11
avatar+108624 
+1

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.

Melody  Jan 7, 2020
edited by Melody  Jan 7, 2020
 #12
avatar+119 
0

Seems you do just the same. I just changed all my votes, your welcome :)

NewMember  Jan 9, 2020

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