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Question-194204

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Question Number 194204 by Rupesh123 last updated on 30/Jun/23
Answered by MM42 last updated on 30/Jun/23
if n=m=0⇒2a_0 −1=(1/2)(2a_0 )⇒a_0 =1 if m=1 , n=0⇒2a_1 −2=(1/2)(a_2 +a_0 ) ⇒6−2=(1/2)(a_2 +1)⇒a_2 =7 m=2 ,n=0⇒2a_2 −3=(1/2)(a_4 +a_0 ) 11=(1/2)(a_4 +1)⇒a_4 =21 m=2 ,n=1⇒a_3 +a_1 −2=(1/2)(a_4 +a_2 ) a_3 +1=(1/2)(21+7)⇒a_3 =13 m=3 , n=0⇒2a_3 −4=(1/2)(a_6 +a_0 ) 44=a_6 +1⇒a_6 =43 m=3 , n=2⇒a_5 +a_1 −2=(1/2)(a_6 +a_4 ) a_5 +1=(1/2)(43+21)⇒a_5 =31 ⇒a_n : 1,3,7,13,21,31,43,... ⇒a_n =a_(n−1) +2n ; n≥1 , a_0 =1 ⇒a_n =1+2+4+...+2n= 1+2(1+2+...+n)=n^2 +n+1 ⇒a_(2024) =2024^2 +2024+1=4098601✓
$${if}\:\:{n}={m}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{0}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{0}} \right)\Rightarrow{a}_{\mathrm{0}} =\mathrm{1} \\ $$$${if}\:\:{m}=\mathrm{1}\:,\:{n}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{1}} −\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{2}} +{a}_{\mathrm{0}} \right) \\ $$$$\Rightarrow\mathrm{6}−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{2}} +\mathrm{1}\right)\Rightarrow{a}_{\mathrm{2}} =\mathrm{7} \\ $$$${m}=\mathrm{2}\:,{n}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{2}} −\mathrm{3}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{4}} +{a}_{\mathrm{0}} \right) \\ $$$$\mathrm{11}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{4}} +\mathrm{1}\right)\Rightarrow{a}_{\mathrm{4}} =\mathrm{21} \\ $$$${m}=\mathrm{2}\:\:,{n}=\mathrm{1}\Rightarrow{a}_{\mathrm{3}} +{a}_{\mathrm{1}} −\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{4}} +{a}_{\mathrm{2}} \right) \\ $$$${a}_{\mathrm{3}} +\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{21}+\mathrm{7}\right)\Rightarrow{a}_{\mathrm{3}} =\mathrm{13} \\ $$$${m}=\mathrm{3}\:,\:{n}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{3}} −\mathrm{4}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{6}} +{a}_{\mathrm{0}} \right) \\ $$$$\mathrm{44}={a}_{\mathrm{6}} +\mathrm{1}\Rightarrow{a}_{\mathrm{6}} =\mathrm{43} \\ $$$${m}=\mathrm{3}\:,\:{n}=\mathrm{2}\Rightarrow{a}_{\mathrm{5}} +{a}_{\mathrm{1}} −\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{6}} +{a}_{\mathrm{4}} \right) \\ $$$${a}_{\mathrm{5}} +\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{43}+\mathrm{21}\right)\Rightarrow{a}_{\mathrm{5}} =\mathrm{31} \\ $$$$\Rightarrow{a}_{{n}} :\:\mathrm{1},\mathrm{3},\mathrm{7},\mathrm{13},\mathrm{21},\mathrm{31},\mathrm{43},… \\ $$$$\Rightarrow{a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{n}\:\:\:;\:{n}\geqslant\mathrm{1}\:\:,\:\:{a}_{\mathrm{0}} =\mathrm{1}\:\:\: \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{1}+\mathrm{2}+\mathrm{4}+…+\mathrm{2}{n}= \\ $$$$\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}+…+{n}\right)={n}^{\mathrm{2}} +{n}+\mathrm{1} \\ $$$$\Rightarrow{a}_{\mathrm{2024}} =\mathrm{2024}^{\mathrm{2}} +\mathrm{2024}+\mathrm{1}=\mathrm{4098601}\checkmark \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 30/Jun/23
Perfect ��

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