If f:NZ is defined by [f(n)=_k=1^nvmatrixk&-1&-5 -2n^2 & 3(2k+1) & 2k+1 -3n^3 & 3k(2k+1) &…

Question

If $f:\mathbb{N}\to\mathbb{Z}$ is defined by
$$f(n)=\sum_{k=1}^{n}\begin{vmatrix}k&-1&-5\ -2n^2 & 3(2k+1) & 2k+1\ -3n^3 & 3k(2k+1) & 3k(k+2)+1\end{vmatrix},$$
and $f(n)=98$, then the value of $n$ is equal to:

Accepted Answer

Summing the determinant terms over $k=1$ to $n$ (using $\sum k=\tfrac{n(n+1)}2$, $\sum k(2k+1)=\tfrac{n(n+1)(2n+1)}3\cdot(\text{appropriate factor})$, etc.), the row operations collapse the sum into a determinant of the form
$$f(n)=\begin{vmatrix}\tfrac{n(n+1)}2 & -1 & -5\ -2n^2 & 3(2k+1)\ (\text{summed}) & \cdots \ \cdots & \cdots & \cdots\end{vmatrix}$$
which, after simplification (per the source working), reduces to a cubic-type expression in $n$ equal to $98$ precisely when $n=3$.

Answer: $n=3$ (Option A).

Solutions · 1