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

 Jan 5, 2020

Best Answer 

 #2
avatar+24344 
+4

Find  \(f(k)\) such that \(\sum \limits_{k=1}^n f(k) = n^3\). :

 

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{k=1}^n f(k)} &=& \mathbf{n^3} \\ \hline \sum \limits_{k=1}^{n+1} f(k) &=& (n+1)^3 \quad | \quad \sum \limits_{k=1}^{n+1} f(k) = \sum \limits_{k=1}^{n} f(k) + f(n+1) \\ \sum \limits_{k=1}^{n} f(k) + f(n+1) &=& (n+1)^3 \quad | \quad \mathbf{\sum \limits_{k=1}^n f(k)=n^3} \\ n^3 + f(n+1) &=& (n+1)^3 \\ f(n+1) &=& (n+1)^3-n^3 \quad | \quad \text{we set } \mathbf{k= n+1} \\ f(k) &=& k^3-\left(k-1\right)^3 \\ \mathbf{f(k)} &=& \mathbf{3k^2-3k+1} \\ \hline \end{array} \)

 

laugh

 Jan 6, 2020
 #1
avatar
-1

From the formulas

 

\(\sum_{k = 1}^n k = \frac{n(n + 1)}{2}, \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n+ 1)}{6}\)

 

it follows that f(k) = k^4/4 + k^3/2 + k^2/4.

 Jan 5, 2020
 #2
avatar+24344 
+4
Best Answer

Find  \(f(k)\) such that \(\sum \limits_{k=1}^n f(k) = n^3\). :

 

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{k=1}^n f(k)} &=& \mathbf{n^3} \\ \hline \sum \limits_{k=1}^{n+1} f(k) &=& (n+1)^3 \quad | \quad \sum \limits_{k=1}^{n+1} f(k) = \sum \limits_{k=1}^{n} f(k) + f(n+1) \\ \sum \limits_{k=1}^{n} f(k) + f(n+1) &=& (n+1)^3 \quad | \quad \mathbf{\sum \limits_{k=1}^n f(k)=n^3} \\ n^3 + f(n+1) &=& (n+1)^3 \\ f(n+1) &=& (n+1)^3-n^3 \quad | \quad \text{we set } \mathbf{k= n+1} \\ f(k) &=& k^3-\left(k-1\right)^3 \\ \mathbf{f(k)} &=& \mathbf{3k^2-3k+1} \\ \hline \end{array} \)

 

laugh

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

That is a really neat solution Heueka !  Thanks.  cool

Melody  Jan 6, 2020
 #5
avatar+24344 
+2

Thank you, Melody !

 

laugh

heureka  Jan 6, 2020
 #4
avatar+108624 
0

I just added tihs  question and answer to our list in the Feature questions sticky topics.    laugh laugh laugh

 

https://web2.0calc.com/questions/feature-questions-1-8th-may19#r23

 

Other people can add impressive threads to this sticky topic too if they want too. :)

 Jan 6, 2020

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