What is the first term in this partial sum $t_{n} = sum_{k=0}^{n} 2^k a_{2^k}$?












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What is the first term in this partial sum $t_{n} = sum_{k=0}^{n} 2^k a_{2^k}$?



I found it to be $a_{1}$(by substituting $ k=0 $ and knowing that 2^0 =1 ) but the book always says that it is $a_{0}$, could anyone remove this discrepancy for me please?










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    $begingroup$
    From the information given in the question, it would seem that you are correct.
    $endgroup$
    – Zachary
    Jan 14 at 14:31










  • $begingroup$
    @Zachary here is the whole information math.stackexchange.com/questions/3071844/…
    $endgroup$
    – hopefully
    Jan 14 at 14:44


















0












$begingroup$


What is the first term in this partial sum $t_{n} = sum_{k=0}^{n} 2^k a_{2^k}$?



I found it to be $a_{1}$(by substituting $ k=0 $ and knowing that 2^0 =1 ) but the book always says that it is $a_{0}$, could anyone remove this discrepancy for me please?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    From the information given in the question, it would seem that you are correct.
    $endgroup$
    – Zachary
    Jan 14 at 14:31










  • $begingroup$
    @Zachary here is the whole information math.stackexchange.com/questions/3071844/…
    $endgroup$
    – hopefully
    Jan 14 at 14:44
















0












0








0





$begingroup$


What is the first term in this partial sum $t_{n} = sum_{k=0}^{n} 2^k a_{2^k}$?



I found it to be $a_{1}$(by substituting $ k=0 $ and knowing that 2^0 =1 ) but the book always says that it is $a_{0}$, could anyone remove this discrepancy for me please?










share|cite|improve this question









$endgroup$




What is the first term in this partial sum $t_{n} = sum_{k=0}^{n} 2^k a_{2^k}$?



I found it to be $a_{1}$(by substituting $ k=0 $ and knowing that 2^0 =1 ) but the book always says that it is $a_{0}$, could anyone remove this discrepancy for me please?







calculus sequences-and-series






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asked Jan 14 at 14:26









hopefullyhopefully

336114




336114








  • 2




    $begingroup$
    From the information given in the question, it would seem that you are correct.
    $endgroup$
    – Zachary
    Jan 14 at 14:31










  • $begingroup$
    @Zachary here is the whole information math.stackexchange.com/questions/3071844/…
    $endgroup$
    – hopefully
    Jan 14 at 14:44
















  • 2




    $begingroup$
    From the information given in the question, it would seem that you are correct.
    $endgroup$
    – Zachary
    Jan 14 at 14:31










  • $begingroup$
    @Zachary here is the whole information math.stackexchange.com/questions/3071844/…
    $endgroup$
    – hopefully
    Jan 14 at 14:44










2




2




$begingroup$
From the information given in the question, it would seem that you are correct.
$endgroup$
– Zachary
Jan 14 at 14:31




$begingroup$
From the information given in the question, it would seem that you are correct.
$endgroup$
– Zachary
Jan 14 at 14:31












$begingroup$
@Zachary here is the whole information math.stackexchange.com/questions/3071844/…
$endgroup$
– hopefully
Jan 14 at 14:44






$begingroup$
@Zachary here is the whole information math.stackexchange.com/questions/3071844/…
$endgroup$
– hopefully
Jan 14 at 14:44












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$begingroup$

You are right. We have $t_n=a_1+2a_2+4a_4+....+2^na_{2^n}$, hence the first term in $t_n$ is $a_1$.






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    1 Answer
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    1 Answer
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    1












    $begingroup$

    You are right. We have $t_n=a_1+2a_2+4a_4+....+2^na_{2^n}$, hence the first term in $t_n$ is $a_1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You are right. We have $t_n=a_1+2a_2+4a_4+....+2^na_{2^n}$, hence the first term in $t_n$ is $a_1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You are right. We have $t_n=a_1+2a_2+4a_4+....+2^na_{2^n}$, hence the first term in $t_n$ is $a_1$.






        share|cite|improve this answer









        $endgroup$



        You are right. We have $t_n=a_1+2a_2+4a_4+....+2^na_{2^n}$, hence the first term in $t_n$ is $a_1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 14:34









        FredFred

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