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A sequence with \(a_1 = 1\) is defined by the recurrence relation \(a_{n+1} = 2^na_n\) for all natural numbers \(n\). If \(a_{23} = 2^p\), then what is \(n\)?

 Jul 24, 2020
 #1
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If I understood your question, then:

 

a=1;b=2^a;c=2^(a+1)*b;printc,", ",;a++;if(a<24, goto1, 0)

 

1, 2, 8 , 32 , 128 , 512 , 2048 , 8192 , 32768 , 131072 , 524288 , 2097152 , 8388608 , 33554432 , 134217728 , 536870912 , 2147483648 , 8589934592 , 3 4359738368 , 13 7438953472 , 54 9755813888 , 219 9023255552 , 8,796,093,022,208, which is the 23rd term = 2^43

 Jul 24, 2020
 #2
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Thanks. That makes sense, but that seems extremely big!

 Jul 24, 2020
 #3
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Can you figure out the 2nd and the 3rd terms, then I will have a better understanding of it?

 Jul 25, 2020
 #4
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I'm sorry I tried but I only got that a0 is 1

HelpBot  Jul 28, 2020
 #5
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I think this is a trick question.

 

\(a_{23}=a_{22+1}\\ so \\n=22\)

 

 

p=23/2*22=253 though

I work it out with the help of an AP sum formula.

 Jul 29, 2020
 #6
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Thanks, that makes sense! But why would you divide 23 by 2??

 Jul 29, 2020
 #7
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It was the end of a fair amount of work.  I worked out the sequence and than used and AP to finish it. 

It was not necessary to answer the question.

Melody  Jul 30, 2020
 #8
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Okay.... thanks

HelpBot  Jul 30, 2020

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