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Sum of a determinant whose entries depend on k or n

This type of question, asked 3 times: 12 Apr 2023, 6 Apr 2024, 5 Apr 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    12 Apr 2023

    86
    Let Dk=∣12k2k−1nn2+n+2n2nn2+nn2+n+2∣D_{k}=\left| \begin{matrix} 1 & 2k & 2k - 1 \\ n & n^{2}+ n + 2 & n^{2} \\ n & n^{2}+ n & n^{2}+ n + 2 \end{matrix} \right|. If ∑k=1nDk=96\sum_{k = 1}^{n} D_{k}= 96, then nn is equal to

  2. Question 2

    6 Apr 2024

    68
    For α,β∈R\alpha,\beta \in R and a natural number nn, let
    Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣. Then 2A10−A8 is A_{r}=\left| \begin{matrix} r & 1 & \frac{n^{2}}{2}+ \alpha \\ 2r & 2 & n^{2}- \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{matrix} \right|.\text{~Then~}2A_{10}-A_{8}\text{~is~}

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  3. Question 3

    5 Apr 2026 S2

    53
    If f:N→Zf:N \rightarrow Z is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣,k∈Nf(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 ∑n=1kf(n)=98\sum_{n= 1}^{k} f(n) = 98, then kk is equal to:

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