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Question-162866

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Question Number 162866 by HongKing last updated on 01/Jan/22
Answered by aleks041103 last updated on 02/Jan/22
determinant ((1,k,k),((2n),(k^2 +k+1),(k^2 +k)),((2n−1),k^2 ,(k^2 +k+1)))= =1. determinant (((k^2 +k+1),(k^2 +k)),(k^2 ,(k^2 +k+1)))−2n determinant ((k,k),(k^2 ,(k^2 +k+1)))+(2n−1) determinant ((k,k),((k^2 +k+1),(k^2 +k)))= determinant (((k^2 +k+1),(k^2 +k)),(k^2 ,(k^2 +k+1)))=(k^2 +k+1)^2 −k^2 (k^2 +k)= =k^4 +k^2 +1+2k^2 +2k^3 +2k−k^4 −k^3 = =k^3 +3k^2 +2k+1 determinant ((k,k),(k^2 ,(k^2 +k+1)))=k(k^2 +k+1)−k^3 =k^2 +k determinant ((k,k),((k^2 +k+1),(k^2 +k)))=k(k^2 +k)−k(k^2 +k+1)= =k(k^2 +k−k^2 −k−1)=−k ⇒U_n =k^3 +3k^2 +2k+1−2n(k^2 +k)−(2n−1)k= ⇒Σ_(n=1) ^k U_n =k(k^3 +3k^2 +2k+1)−k^2 (k+1)^2 +k−k^2 (k+1)= =k^4 +3k^3 +2k^2 +k−(k^4 +2k^3 +k^2 )+k−k^3 −k^2 = =k^3 +k^2 +2k−k^3 −k^2 =2k ⇒2k=72⇒k=36
$$\begin{vmatrix}{\mathrm{1}}&{{k}}&{{k}}\\{\mathrm{2}{n}}&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\\{\mathrm{2}{n}−\mathrm{1}}&{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}= \\ $$$$=\mathrm{1}.\begin{vmatrix}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\\{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}−\mathrm{2}{n}\begin{vmatrix}{{k}}&{{k}}\\{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}+\left(\mathrm{2}{n}−\mathrm{1}\right)\begin{vmatrix}{{k}}&{{k}}\\{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\end{vmatrix}= \\ $$$$\begin{vmatrix}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\\{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}=\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \left({k}^{\mathrm{2}} +{k}\right)= \\ $$$$={k}^{\mathrm{4}} +{k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}^{\mathrm{3}} +\mathrm{2}{k}−{k}^{\mathrm{4}} −{k}^{\mathrm{3}} = \\ $$$$={k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1} \\ $$$$\begin{vmatrix}{{k}}&{{k}}\\{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}={k}\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)−{k}^{\mathrm{3}} ={k}^{\mathrm{2}} +{k} \\ $$$$\begin{vmatrix}{{k}}&{{k}}\\{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\end{vmatrix}={k}\left({k}^{\mathrm{2}} +{k}\right)−{k}\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)= \\ $$$$={k}\left({k}^{\mathrm{2}} +{k}−{k}^{\mathrm{2}} −{k}−\mathrm{1}\right)=−{k} \\ $$$$\Rightarrow{U}_{{n}} ={k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}−\mathrm{2}{n}\left({k}^{\mathrm{2}} +{k}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right){k}= \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{U}_{{n}} ={k}\left({k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)−{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} +{k}−{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)= \\ $$$$={k}^{\mathrm{4}} +\mathrm{3}{k}^{\mathrm{3}} +\mathrm{2}{k}^{\mathrm{2}} +{k}−\left({k}^{\mathrm{4}} +\mathrm{2}{k}^{\mathrm{3}} +{k}^{\mathrm{2}} \right)+{k}−{k}^{\mathrm{3}} −{k}^{\mathrm{2}} = \\ $$$$={k}^{\mathrm{3}} +{k}^{\mathrm{2}} +\mathrm{2}{k}−{k}^{\mathrm{3}} −{k}^{\mathrm{2}} =\mathrm{2}{k} \\ $$$$\Rightarrow\mathrm{2}{k}=\mathrm{72}\Rightarrow{k}=\mathrm{36} \\ $$
Commented by HongKing last updated on 02/Jan/22
perfect my dear Sir thank you so much
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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