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

Jan 5, 2020

#2
+4

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

$$\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{k=1}^n f(k)} &=& \mathbf{n^3} \\ \hline \sum \limits_{k=1}^{n+1} f(k) &=& (n+1)^3 \quad | \quad \sum \limits_{k=1}^{n+1} f(k) = \sum \limits_{k=1}^{n} f(k) + f(n+1) \\ \sum \limits_{k=1}^{n} f(k) + f(n+1) &=& (n+1)^3 \quad | \quad \mathbf{\sum \limits_{k=1}^n f(k)=n^3} \\ n^3 + f(n+1) &=& (n+1)^3 \\ f(n+1) &=& (n+1)^3-n^3 \quad | \quad \text{we set } \mathbf{k= n+1} \\ f(k) &=& k^3-\left(k-1\right)^3 \\ \mathbf{f(k)} &=& \mathbf{3k^2-3k+1} \\ \hline \end{array}$$ Jan 6, 2020

#1
-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
+4

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

$$\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{k=1}^n f(k)} &=& \mathbf{n^3} \\ \hline \sum \limits_{k=1}^{n+1} f(k) &=& (n+1)^3 \quad | \quad \sum \limits_{k=1}^{n+1} f(k) = \sum \limits_{k=1}^{n} f(k) + f(n+1) \\ \sum \limits_{k=1}^{n} f(k) + f(n+1) &=& (n+1)^3 \quad | \quad \mathbf{\sum \limits_{k=1}^n f(k)=n^3} \\ n^3 + f(n+1) &=& (n+1)^3 \\ f(n+1) &=& (n+1)^3-n^3 \quad | \quad \text{we set } \mathbf{k= n+1} \\ f(k) &=& k^3-\left(k-1\right)^3 \\ \mathbf{f(k)} &=& \mathbf{3k^2-3k+1} \\ \hline \end{array}$$ heureka Jan 6, 2020
#3
+1

That is a really neat solution Heueka !  Thanks. Melody  Jan 6, 2020
#5
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Thank you, Melody ! heureka  Jan 6, 2020
#4
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I just added tihs  question and answer to our list in the Feature questions sticky topics.   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