Question
If $f:N \rightarrow Z$ is defined by
$f(n) = \left| \begin{matrix} n & - 1 & - 5 \\ - 2n^{2} & 3(2k + 1) & 2k + 1 \\ - 3n^{3} & 3k(2k + 1) & 3k(k + 2) + 1 \end{matrix} \right|,k \in N$, and $\sum_{n = 1}^{k}\mspace{2mu} f(n) = 98$, then $k$ is equal to:
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Solution
Sol. [Key Concept: Expand the determinant, use $\sum\ n = \frac{k(k + 1)}{2}$, $\sum\ n^{2}$, $\sum\ n^{3}$, set the total to $98$ and solve for $k$.]
$$f(n) = \left| \begin{matrix} n & - 1 & - 5 \\ - 2n^{2} & 3(2k + 1) & 2k + 1 \\ - 3n^{3} & 3k(2k + 1) & 3k(k + 2) + 1 \end{matrix} \right|$$
$$ {\sum_{n = 1}^{k}\mspace{2mu} f(n) = \left| \begin{matrix} \frac{k(k + 1)}{2} & - 1 & - 5 \\ - 2\frac{k(k + 1)(2k + 1)}{6} & 3(2k + 1) & 2k + 1 \\ - 3\frac{k^{2}(k + 1)^{2}}{4} & 3k(2k + 1) & 3k(k + 2) + 1 \end{matrix} \right| = 98}$$
$$ {\Rightarrow \frac{k(k + 1)(2k + 1)}{2} \cdot \frac{7}{3} = 98}$$
$$ {\Rightarrow k(k + 1)(2k + 1) = 84}$$
$$ {\Rightarrow k = 3 }$$
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