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Question Number 162866 by HongKing last updated on 01/Jan/22

Answered by aleks041103 last updated on 02/Jan/22

$$\left|\begin{array}{ccc}1& k& k\\ 2n& k^2+k+1& k^2+k\\ 2n-1& k^2& k^2+k+1\end{array}\right|=$$$$=1.\left|\begin{array}{cc}{k}^2+k+1& k^2+k\\ k^2& k^2+k+1\end{array}\right|-2n\left|\begin{array}{cc}k& k\\ k^2& k^2+k+1\end{array}\right|+(2n-1)\left|\begin{array}{cc}k& k\\ k^2+k+1& k^2+k\end{array}\right|=$$$$\left|\begin{array}{cc}{k}^2+k+1& k^2+k\\ k^2& k^2+k+1\end{array}\right|={(k^2+k+1)}^2-k^2(k^2+k)=$$$$=k^4+k^2+1+2k^2+2k^3+2k-k^4-k^3=$$$$=k^3+3k^2+2k+1$$$$\left|\begin{array}{cc}k& k\\ k^2& k^2+k+1\end{array}\right|=k(k^2+k+1)-k^3=k^2+k$$$$\left|\begin{array}{cc}k& k\\ k^2+k+1& k^2+k\end{array}\right|=k(k^2+k)-k(k^2+k+1)=$$$$=k(k^2+k-k^2-k-1)=-k$$$$\Rightarrow U_n=k^3+3k^2+2k+1-2n(k^2+k)-(2n-1)k=$$$$\Rightarrow \underset{n=1}{\sum ^k}{U}_n=k(k^3+3k^2+2k+1)-k^2{(k+1)}^2+k-k^2(k+1)=$$$$=k^4+3k^3+2k^2+k-(k^4+2k^3+k^2)+k-k^3-k^2=$$$$=k^3+k^2+2k-k^3-k^2=2k$$$$\Rightarrow 2k=72\Rightarrow k=36$$

Commented by HongKing last updated on 02/Jan/22

$$\mathrm{perfect}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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