Question Number 926 by 123456 last updated on 26/Apr/15
$${a}\left({n},{m}\right)=\begin{cases}{{m}}&{{n}\leqslant\mathrm{0}}\\{{na}\left({n}+{m}−\mathrm{1},{m}\right)+{m}}&{{n}>\mathrm{0}\vee{m}\leqslant\mathrm{0}}\\{{na}\left({n}−{m},{m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} }&{{n}>\mathrm{0}\vee{m}>\mathrm{0}}\end{cases} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)=??? \\ $$$${a}\left(\mathrm{10},\mathrm{10}\right)=?? \\ $$
Commented by 123456 last updated on 27/Apr/15
$${a}\left(\mathrm{0},{m}\right)={m} \\ $$
Commented by 123456 last updated on 27/Apr/15
$${a}\left(\mathrm{1},{m}\right)=\begin{cases}{\mathrm{2}{m}}&{{m}\leqslant\mathrm{0}}\\{\mathrm{4}{m}−\mathrm{2}}&{\mathrm{0}<{m}<\mathrm{1}}\\{{m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{1}}&{{m}\geqslant\mathrm{1}}\end{cases} \\ $$$${m}\leqslant\mathrm{0} \\ $$$${a}\left(\mathrm{1},{m}\right)={a}\left({m},{m}\right)+{m}={m}+{m}=\mathrm{2}{m} \\ $$$${m}>\mathrm{0} \\ $$$${a}\left(\mathrm{1},{m}\right)={a}\left(\mathrm{1}−{m},{m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} \\ $$$${m}\geqslant\mathrm{1} \\ $$$${a}\left(\mathrm{1}−{m},{m}−\mathrm{1}\right)={m}−\mathrm{1} \\ $$$${a}\left(\mathrm{1},{m}\right)={m}−\mathrm{1}+{m}+{m}^{\mathrm{2}} ={m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{1} \\ $$$$\mathrm{0}<{m}<\mathrm{1} \\ $$$${a}\left(\mathrm{1}−{m},{m}−\mathrm{1}\right)=\left(\mathrm{1}−{m}\right){a}\left(−\mathrm{1},{m}−\mathrm{1}\right)+{m}−\mathrm{1} \\ $$$$=\left(\mathrm{1}−{m}\right)\left({m}−\mathrm{1}\right)+{m}−\mathrm{1} \\ $$$$=−\left({m}−\mathrm{1}\right)^{\mathrm{2}} +{m}−\mathrm{1} \\ $$$$=−\left({m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}\right)+{m}−\mathrm{1} \\ $$$$=−{m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{1}+{m}−\mathrm{1} \\ $$$$=−{m}^{\mathrm{2}} +\mathrm{3}{m}−\mathrm{2} \\ $$$${a}\left(\mathrm{1},{m}\right)=−{m}^{\mathrm{2}} +\mathrm{3}{m}−\mathrm{2}+{m}+{m}^{\mathrm{2}} =\mathrm{4}{m}−\mathrm{2} \\ $$
Commented by 123456 last updated on 27/Apr/15
$${a}\left({m},{m}\right)=\begin{cases}{{m}}&{{m}\leqslant\mathrm{0}}\\{\mathrm{2}{m}^{\mathrm{2}} }&{{m}>\mathrm{0}}\end{cases} \\ $$$${m}\leqslant\mathrm{0} \\ $$$${a}\left({m},{m}\right)={m} \\ $$$${m}>\mathrm{0} \\ $$$${a}\left({m},{m}\right)={ma}\left(\mathrm{0},{m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} \\ $$$${a}\left(\mathrm{0},{m}−\mathrm{1}\right)={m}−\mathrm{1} \\ $$$${a}\left({m},{m}\right)={m}\left({m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} \\ $$$$={m}^{\mathrm{2}} −{m}+{m}+{m}^{\mathrm{2}} \\ $$$$=\mathrm{2}{m}^{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 28/Apr/15
$${a}\left(\mathrm{5},\mathrm{5}\right)=\mathrm{5}{a}\left(\mathrm{0},\mathrm{4}\right)+\mathrm{5}+\mathrm{5}^{\mathrm{2}} \\ $$$$=\mathrm{5}×\mathrm{4}+\mathrm{5}+\mathrm{25}=\mathrm{50} \\ $$$$\mathrm{a}\left(\mathrm{10},\mathrm{10}\right)=\mathrm{10}×\mathrm{9}+\mathrm{10}+\mathrm{10}^{\mathrm{2}} =\mathrm{200} \\ $$