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

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