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Question Number 194204 by Rupesh123 last updated on 30/Jun/23
Answered by MM42 last updated on 30/Jun/23
ifn=m=0⇒2a0−1=12(2a0)⇒a0=1ifm=1,n=0⇒2a1−2=12(a2+a0)⇒6−2=12(a2+1)⇒a2=7m=2,n=0⇒2a2−3=12(a4+a0)11=12(a4+1)⇒a4=21m=2,n=1⇒a3+a1−2=12(a4+a2)a3+1=12(21+7)⇒a3=13m=3,n=0⇒2a3−4=12(a6+a0)44=a6+1⇒a6=43m=3,n=2⇒a5+a1−2=12(a6+a4)a5+1=12(43+21)⇒a5=31⇒an:1,3,7,13,21,31,43,...⇒an=an−1+2n;n⩾1,a0=1⇒an=1+2+4+...+2n=1+2(1+2+...+n)=n2+n+1⇒a2024=20242+2024+1=4098601✓
Commented by Rupesh123 last updated on 30/Jun/23
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