Question Number 25482 by math solver last updated on 11/Dec/17
Commented by prakash jain last updated on 11/Dec/17
$${x}=\frac{\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{3}−\mathrm{4}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\pm{i}}{\mathrm{2}} \\ $$$${x}=\frac{\sqrt{\mathrm{3}}+{i}}{\mathrm{2}},\:\omega=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{x}=−{i}\omega \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\left(−{iw}\right)^{{n}} −\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\left(−\frac{\mathrm{1}}{{iw}}\right)^{{n}} \\ $$$$\frac{−{iw}\left(\left(−{iw}\right)^{\mathrm{24}} −\mathrm{1}\right)}{−{iw}−\mathrm{1}}−\frac{−\frac{\mathrm{1}}{{iw}}\left(\frac{\mathrm{1}}{\left(−{iw}\right)^{\mathrm{24}} }−\mathrm{1}\right)}{−\frac{\mathrm{1}}{{iw}}−\mathrm{1}}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{recheck}. \\ $$
Commented by prakash jain last updated on 11/Dec/17
Did u find any mistake in my work. I will also check later
Commented by prakash jain last updated on 12/Dec/17
0 is the correct answer.
Answered by ajfour last updated on 12/Dec/17
$$\mathrm{1}+{x}^{\mathrm{2}} ={x}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\left({x}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\pm\frac{{i}}{\mathrm{2}}\:={e}^{\pm\pi/\mathrm{3}} ={i}\omega^{\mathrm{2}} ,\:−{i}\omega \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\left({x}^{{n}} −\frac{\mathrm{1}}{{x}^{{n}} }\right)\:=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{24}} \right)}{\mathrm{1}−{x}}−\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{24}} }}{{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\:\:\:=\frac{{x}−{x}^{\mathrm{25}} }{\mathrm{1}−{x}}\:−\:\frac{\mathrm{1}−{x}^{\mathrm{24}} }{{x}^{\mathrm{24}} \left(\mathrm{1}−{x}\right)} \\ $$$$\:\:=\frac{{x}^{\mathrm{25}} −{x}^{\mathrm{49}} −\mathrm{1}+{x}^{\mathrm{24}} }{{x}^{\mathrm{24}} \left(\mathrm{1}−{x}\right)\:} \\ $$$$\:\:{for}\:{x}=−{i}\omega \\ $$$$\:\:{S}=\frac{−{i}\omega+{i}\omega−\mathrm{1}+\mathrm{1}}{\left(\mathrm{1}+{i}\omega\right)}\:=\mathrm{0} \\ $$$${for}\:{x}={i}\omega^{\mathrm{2}} \\ $$$$\:\:\:{S}=\frac{{i}\omega^{\mathrm{2}} −{i}\omega^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\left(\mathrm{1}+{i}\omega^{\mathrm{2}} \right)}\:=\mathrm{0}\:. \\ $$
Commented by prakash jain last updated on 13/Dec/17
$$\mathrm{Thanks}. \\ $$