Question Number 62322 by aliesam last updated on 19/Jun/19 | ||
Commented by arcana last updated on 19/Jun/19 | ||
$$\beta_{\mathrm{1}} =\alpha+\alpha^{\mathrm{6}} \\ $$$$\beta_{\mathrm{2}} =\left(\alpha+\alpha^{\mathrm{6}} \right)^{\mathrm{2}} −\mathrm{2}=\alpha^{\mathrm{2}} +\mathrm{2}\alpha^{\mathrm{7}} +\alpha^{\mathrm{12}} −\mathrm{2} \\ $$$$\beta_{\mathrm{2}} =\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} \\ $$$$\beta_{\mathrm{3}} =\left(\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} \right)^{\mathrm{2}} −\mathrm{2}=\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{7}} +\alpha^{\mathrm{10}} −\mathrm{2} \\ $$$$\beta_{\mathrm{3}} =\alpha^{\mathrm{4}} +\alpha^{\mathrm{3}} \\ $$$$\beta_{\mathrm{4}} =\left(\alpha^{\mathrm{4}} +\alpha^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}=\alpha^{\mathrm{8}} +\mathrm{2}\alpha^{\mathrm{7}} +\alpha^{\mathrm{6}} −\mathrm{2} \\ $$$$\beta_{\mathrm{4}} =\alpha+\alpha^{\mathrm{6}} =\beta_{\mathrm{1}} \\ $$$$\Rightarrow\underset{\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\beta_{{k}} =\mathrm{33}\underset{\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\beta_{{k}} =\mathrm{33}\left[\beta_{\mathrm{1}} +\beta_{\mathrm{2}} +\beta_{\mathrm{3}} \right] \\ $$$$=\mathrm{33}\left[\alpha+\alpha^{\mathrm{6}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{3}} \right]=−\mathrm{33} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by alphaprime last updated on 19/Jun/19 | ||
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