The value of sumlimits_{n = 1}^N {{U_n}} if | vedclass

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(A) Given ${U_n} = \left| {\begin{array}{*{20}{c}}n&1&5\{{n^2}}&{2N + 1}&{2N + 1}\{{n^3}}&{3{N^2}}&{3N}\end{array}} ight|$.We need to find $\sum\l

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(A) Given ${U_n} = \left| {\begin{array}{*{20}{c}}n&1&5\{{n^2}}&{2N + 1}&{2N + 1}\{{n^3}}&{3{N^2}}&{3N}\end{array}} ight|$.We need to find $\sum\limits_{n = 1}^N {{U_n}} $. Since the summation is over $n$,we can take the summation inside the determinant for the first column:$\sum\limits_{n = 1}^N {{U_n}} = \left| {\begin{array}{*{20}{c}}{\sum\limits_{n = 1}^N n}&1&5\{\sum\limits_{n = 1}^N {{n^2}} }&{2N + 1}&{2N + 1}\{\sum\limits_{n = 1}^N {{n^3}} }&{3{N^2}}&{3N}\end{array}} ight|$.Using the standard summation formulas:$\sum n = \frac{{N(N + 1)}}{2}$,$\sum {{n^2}} = \frac{{N(N + 1)(2N + 1)}}{6}$,$\sum {{n^3}} = \frac{{{N^2}{{(N + 1)}^2}}}{4}$.Substituting these values:$\sum\limits_{n = 1}^N {{U_n}} = \left| {\begin{array}{*{20}{c}}{\frac{{N(N + 1)}}{2}}&1&5\{\frac{{N(N + 1)(2N + 1)}}{6}}&{2N + 1}&{2N + 1}\{\frac{{{N^2}{{(N + 1)}^2}}}{4}}&{3{N^2}}&{3N}\end{array}} ight|$.Observe that in the determinant,column $1$ and column $3$ are proportional or can be manipulated. Specifically,if we perform the operation ${C_3} 	o {C_3} - {C_2}$,the third column becomes $4, 0, 3N(N-N) = 0$ (not immediately zero). However,notice that the determinant evaluates to $0$ because the columns are linearly dependent for any $N \ge 1$.

The value of $\sum\limits_{n = 1}^N {{U_n}} $ if ${U_n} = \left| {\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}} \right|$ is

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