Question Number 440 by 123456 last updated on 25/Jan/15
$${a}\left({n}+\mathrm{1}\right)=\left[{a}\left({n}\right)+\mathrm{1}\right]\mathrm{cos}\left(\frac{\pi{n}}{\mathrm{2}}\right)+\left[{a}\left({n}−\mathrm{1}\right)+{n}\right]\mathrm{sin}\:\left(\frac{\pi{n}}{\mathrm{2}}\right) \\ $$$${a}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${a}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${a}\left(\mathrm{10}\right)=? \\ $$
Answered by prakash jain last updated on 04/Jan/15
$${a}\left(\mathrm{2}\right)=\left[{a}\left(\mathrm{0}\right)+\mathrm{1}\right]\mathrm{sin}\:\frac{\pi}{\mathrm{2}}=\mathrm{1} \\ $$$${a}\left(\mathrm{3}\right)=\left[{a}\left(\mathrm{2}\right)+\mathrm{1}\right]\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{2}}=−\mathrm{2} \\ $$$${a}\left(\mathrm{4}\right)=\left[{a}\left(\mathrm{2}\right)+\mathrm{3}\right]\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}}=−\mathrm{4} \\ $$$${a}\left(\mathrm{5}\right)=\left[{a}\left(\mathrm{4}\right)+\mathrm{1}\right]\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{2}}=−\mathrm{3} \\ $$$${a}\left(\mathrm{6}\right)=\left[{a}\left(\mathrm{4}\right)+\mathrm{5}\right]\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{2}}=\mathrm{1} \\ $$$${a}\left(\mathrm{7}\right)=\left[{a}\left(\mathrm{6}\right)+\mathrm{1}\right]\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{2}}=−\mathrm{2} \\ $$$${a}\left(\mathrm{8}\right)=\left[{a}\left(\mathrm{6}\right)+\mathrm{7}\right]\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{2}}=−\mathrm{5} \\ $$$${a}\left(\mathrm{9}\right)=\left[{a}\left(\mathrm{8}\right)+\mathrm{1}\right]\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{2}}=−\mathrm{4} \\ $$$${a}\left(\mathrm{10}\right)=\left[{a}\left(\mathrm{8}\right)+\mathrm{9}\right]\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{2}}=\mathrm{4} \\ $$