If Dr=|rxn(n+1)22r−1yn23r−2zn(3n−1)2| , then ∑r=1nDr=

Dr=|rxn(n+1)22r−1yn23r−2zn(3n−1)2| ΣDr=|ΣrΣxΣn(n+1)2Σ(2r−1)ΣyΣn2Σ(3r−2)ΣzΣn(3n−1)2| ∑r=1nr=1+2+3+⋯+n=n(n+1)2 ∑r=1n2r−1=∑r=1n2r−∑r=1n1=2∑r=1nr−∑r=1n1=n(n+1)−n=n2 ∑r=1n(3r−2)=3∑r=1nr−2∑r=1n1=3n(n+1)2−2n =3n2+3n−4n2 =3n2−n2 =n2(3n−1) Σx=nx;Σy=ny,Σz=nz ∑n(n+1)2=n(n+1)2Σ1=n2(n+1)2 Σn2=n3 ∑n(3n−1)2=n2(3n−1)2 Dr=|n(n+1)2nxn2(n+1)2n2nyn3n2(3n−1)nzn2(3n−1)2|=n|n(n+1)2nxn(n+1)2n2nyn2n(3n−1)2n2n(3n−1)2| When two rows/ columns are identical then determinant is zero.

If Dr=|rxn(n+1)22r−1yn23r−2zn(3n−1)2| , then ∑r=1nDr=

Dr=|rxn(n+1)22r−1yn23r−2zn(3n−1)2| ΣDr=|ΣrΣxΣn(n+1)2Σ(2r−1)ΣyΣn2Σ(3r−2)ΣzΣn(3n−1)2| ∑r=1nr=1+2+3+⋯+n=n(n+1)2 ∑r=1n2r−1=∑r=1n2r−∑r=1n1=2∑r=1nr−∑r=1n1=n(n+1)−n=n2 ∑r=1n(3r−2)=3∑r=1nr−2∑r=1n1=3n(n+1)2−2n =3n2+3n−4n2 =3n2−n2 =n2(3n−1) Σx=nx;Σy=ny,Σz=nz ∑n(n+1)2=n(n+1)2Σ1=n2(n+1)2 Σn2=n3 ∑n(3n−1)2=n2(3n−1)2 Dr=|n(n+1)2nxn2(n+1)2n2nyn3n2(3n−1)nzn2(3n−1)2|=n|n(n+1)2nxn(n+1)2n2nyn2n(3n−1)2n2n(3n−1)2| When two rows/ columns are identical then determinant is zero.

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If $D_{r} = | \begin{matrix} r & x & \frac{n (n + 1)}{2} \\ 2 r - 1 & y & n^{2} \\ 3 r - 2 & z & \frac{n (3 n - 1)}{2} \end{matrix} |$ , then $\sum_{r = 1}^{n} D_{r} =$

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Detailed Solution

D_{r} = | \begin{matrix} r & x & \frac{n (n + 1)}{2} \\ 2 r - 1 & y & n^{2} \\ 3 r - 2 & z & \frac{n (3 n - 1)}{2} \end{matrix} |

$Σ D_{r} = | \begin{matrix} Σ r & Σ x & Σ \frac{n (n + 1)}{2} \\ Σ (2 r - 1) & Σ y & Σ n^{2} \\ Σ (3 r - 2) & Σ z & Σ \frac{n (3 n - 1)}{2} \end{matrix} |$

$\sum_{r = 1}^{n} r = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$

$\sum_{r = 1}^{n} 2 r - 1 = \sum_{r = 1}^{n} 2 r - \sum_{r = 1}^{n} 1 = 2 \sum_{r = 1}^{n} r - \sum_{r = 1}^{n} 1 = n (n + 1) - n = n^{2}$

$\sum_{r = 1}^{n} (3 r - 2) = 3 \sum_{r = 1}^{n} r - 2 \sum_{r = 1}^{n} 1 = \frac{3 n (n + 1)}{2} - 2 n$

$= \frac{3 n^{2} + 3 n - 4 n}{2}$

$= \frac{3 n^{2} - n}{2}$

$= \frac{n}{2} (3 n - 1)$

$Σ x = n x; Σ y = n y, Σ z = n z$

$\sum \frac{n (n + 1)}{2} = \frac{n (n + 1)}{2} Σ 1 = \frac{n^{2} (n + 1)}{2}$

$Σ n^{2} = n^{3}$

$\sum \frac{n (3 n - 1)}{2} = \frac{n^{2} (3 n - 1)}{2}$

$D_{r} = | \begin{matrix} \frac{n (n + 1)}{2} & n x & \frac{n^{2} (n + 1)}{2} \\ n^{2} & n y & n^{3} \\ \frac{n}{2} (3 n - 1) & n z & \frac{n^{2} (3 n - 1)}{2} \end{matrix} | = n | \begin{matrix} \frac{n (n + 1)}{2} & n x & \frac{n (n + 1)}{2} \\ n^{2} & n y & n^{2} \\ \frac{n (3 n - 1)}{2} & n^{2} & \frac{n (3 n - 1)}{2} \end{matrix} |$