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

 Aug 14, 2019
 #1
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We know that f(k) is a quadratic function in k.

Let f(k) = ak2 + bk + c.

\(\phantom{=}\displaystyle \sum_{k = 1}^n f(k)\\ = a \displaystyle\sum_{k=1}^n k^2 + b\displaystyle\sum_{k=1}^n k + c\displaystyle\sum_{k=1}^n 1\\ = \dfrac{an(n + 1)(2n + 1)}{6} + \dfrac{bn(n + 1)}{2} + cn\\ = \dfrac{n}{6} \left(a(2n^2 + 3n + 1) + 3b(n + 1) + 6c\right)\\ = \dfrac{n(2an^2 + (3a + 3b)n + a + 3b + 6c)}{6}\\ \text{Compare with }n^3,\\ n^3 = \dfrac{n(2an^2 + (3a + 3b)n + a + 3b + 6c)}{6}\\ 6n^3 = 2an^3 + 3(a + b)n^2 + (a + 3b + 6c)n\\ \text{Comparing coefficients, }\\ 2a = 6 \implies a = 3\\ 3(3 + b) = 0 \implies b = -3\\ 3 + 3(-3) + 6c = 0 \implies c= 1\\ \therefore f(k) = 3k^2 -3k + 1\)

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 Aug 14, 2019
 #2
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\(f(n)=\sum_{k=1}^{n} f(k) -\sum_{k=1}^{n-1} f(k)=n^3-(n-1)^3=3n^2-3n+1\)

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 Aug 14, 2019

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