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?
calculus sequences-and-series
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add a comment |
$begingroup$
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?
calculus sequences-and-series
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2
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From the information given in the question, it would seem that you are correct.
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– Zachary
Jan 14 at 14:31
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@Zachary here is the whole information math.stackexchange.com/questions/3071844/…
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– hopefully
Jan 14 at 14:44
add a comment |
$begingroup$
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?
calculus sequences-and-series
$endgroup$
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
calculus sequences-and-series
asked Jan 14 at 14:26
hopefullyhopefully
336114
336114
2
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From the information given in the question, it would seem that you are correct.
$endgroup$
– Zachary
Jan 14 at 14:31
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@Zachary here is the whole information math.stackexchange.com/questions/3071844/…
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– hopefully
Jan 14 at 14:44
add a comment |
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
add a comment |
1 Answer
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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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$begingroup$
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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add a comment |
$begingroup$
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 $a_1$.
$endgroup$
add a comment |
$begingroup$
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 $a_1$.
$endgroup$
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 $a_1$.
answered Jan 14 at 14:34
FredFred
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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/…
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– hopefully
Jan 14 at 14:44