If $$Di=1nn2in2+n+1n2+n2i-1n2n2+n+1and$$∑$$i=1nDi=56then$$ n must be $$=$$ $$______$$

$$Di=1nn2in2+n+1n2+n2i-1n2n2+n+1Write$$ in to $$n-determinants$$ and apply∑$$i=1nDi=$$∑$$i=1n1nn$$∑$$i=1n2in2+n+1n2+n$$∑$$i=1n2i-1n2n2+n+1$$ ∑$$i=1nDi=nnnn(n+1)n2+n+1n2+nn2n2n2+n+1$$ ∑$$i=1nDi=n00n2+n10n20n+1=n(n+1)$$⇒$$n(n+1)=56$$∴$$n=7$$

Questions

If $D_{i =} |\begin{matrix} 1 & n & n \\ 2 i & n^{2} + n + 1 & n^{2} + n \\ 2 i - 1 & n^{2} & n^{2} + n + 1 \end{matrix}| and$ $\sum_{i = 1}^{n} D_{i} = 56$ then n must be = ______

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detailed solution

$D_{i =} |\begin{matrix} 1 & n & n \\ 2 i & n^{2} + n + 1 & n^{2} + n \\ 2 i - 1 & n^{2} & n^{2} + n + 1 \end{matrix}|$ Write in to n-determinants and apply

$\sum_{i = 1}^{n} D_{i} = |\begin{matrix} \sum_{i = 1}^{n} 1 & n & n \\ \sum_{i = 1}^{n} 2 i & n^{2} + n + 1 & n^{2} + n \\ \sum_{i = 1}^{n} 2 i - 1 & n^{2} & n^{2} + n + 1 \end{matrix}| \sum_{i = 1}^{n} D_{i} = |\begin{matrix} n & n & n \\ n (n + 1) & n^{2} + n + 1 & n^{2} + n \\ n^{2} & n^{2} & n^{2} + n + 1 \end{matrix}| \sum_{i = 1}^{n} D_{i} = |\begin{matrix} n & 0 & 0 \\ n^{2} + n & 1 & 0 \\ n^{2} & 0 & n + 1 \end{matrix}| = n (n + 1) \Rightarrow n (n + 1) = 56 ∴ n = 7$