Cannot find mistake in exponentiation when trying to simplify a general sequence












0












$begingroup$


I am dealing with a general sequence, $a_n$, for which $S_n$ is the sequence of its partial sums.

I know that $a_n = S_{n} - S_{n-1}$, and therefore $a_n = S_{n+1} - S_{n}$
also.



I want to find $a_n$ given:
$$ S_n = k - frac{1}{3^{n+1}} $$



Here is the correct, confirmed calculation:
$$ a_n = S_{n} - S_{n-1} = k - frac{1}{3^{n+1}} - k +frac{1}{3^n}$$
$$ a_n = frac{1}{3^n} - frac{1}{3^{n+1}}$$
$$ a_n = frac{1-3^{-1}}{3^n} $$
$$ a_n = frac{2/3}{3^n} $$
$$ boxed{a_n = frac{2}{3^{n+1}}}$$



Fine and dandy, but when I try to solve it from a different angle, I get a wrong answer and can't understand where have I went wrong:



$$ a_n = S_{n+1} - S_{n} = k - frac{1}{3^{n+2}} - k +frac{1}{3^{n+1}}$$
$$ a_n = frac{1}{3^{n+1}} - frac{1}{3^{n+2}}$$
$$ a_n = frac{3^{n+2} - 3^{n+1}}{3^{n+1}cdot 3^{n+2}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n+3}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n}(3^3)} $$
$$ a_n = frac{3^{n - 2n}(3^2 - 3^1)}{3^3} = frac{3^{-n}(9 - 3)}{27} = frac{6}{27cdot 3^n} = frac{2}{9cdot 3^n} boxed{neq frac{2}{3^{n+2}}}$$



Help is greatly appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
    $endgroup$
    – Gerhard S.
    Dec 22 '18 at 17:05






  • 1




    $begingroup$
    Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
    $endgroup$
    – timtfj
    Dec 22 '18 at 17:20
















0












$begingroup$


I am dealing with a general sequence, $a_n$, for which $S_n$ is the sequence of its partial sums.

I know that $a_n = S_{n} - S_{n-1}$, and therefore $a_n = S_{n+1} - S_{n}$
also.



I want to find $a_n$ given:
$$ S_n = k - frac{1}{3^{n+1}} $$



Here is the correct, confirmed calculation:
$$ a_n = S_{n} - S_{n-1} = k - frac{1}{3^{n+1}} - k +frac{1}{3^n}$$
$$ a_n = frac{1}{3^n} - frac{1}{3^{n+1}}$$
$$ a_n = frac{1-3^{-1}}{3^n} $$
$$ a_n = frac{2/3}{3^n} $$
$$ boxed{a_n = frac{2}{3^{n+1}}}$$



Fine and dandy, but when I try to solve it from a different angle, I get a wrong answer and can't understand where have I went wrong:



$$ a_n = S_{n+1} - S_{n} = k - frac{1}{3^{n+2}} - k +frac{1}{3^{n+1}}$$
$$ a_n = frac{1}{3^{n+1}} - frac{1}{3^{n+2}}$$
$$ a_n = frac{3^{n+2} - 3^{n+1}}{3^{n+1}cdot 3^{n+2}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n+3}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n}(3^3)} $$
$$ a_n = frac{3^{n - 2n}(3^2 - 3^1)}{3^3} = frac{3^{-n}(9 - 3)}{27} = frac{6}{27cdot 3^n} = frac{2}{9cdot 3^n} boxed{neq frac{2}{3^{n+2}}}$$



Help is greatly appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
    $endgroup$
    – Gerhard S.
    Dec 22 '18 at 17:05






  • 1




    $begingroup$
    Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
    $endgroup$
    – timtfj
    Dec 22 '18 at 17:20














0












0








0





$begingroup$


I am dealing with a general sequence, $a_n$, for which $S_n$ is the sequence of its partial sums.

I know that $a_n = S_{n} - S_{n-1}$, and therefore $a_n = S_{n+1} - S_{n}$
also.



I want to find $a_n$ given:
$$ S_n = k - frac{1}{3^{n+1}} $$



Here is the correct, confirmed calculation:
$$ a_n = S_{n} - S_{n-1} = k - frac{1}{3^{n+1}} - k +frac{1}{3^n}$$
$$ a_n = frac{1}{3^n} - frac{1}{3^{n+1}}$$
$$ a_n = frac{1-3^{-1}}{3^n} $$
$$ a_n = frac{2/3}{3^n} $$
$$ boxed{a_n = frac{2}{3^{n+1}}}$$



Fine and dandy, but when I try to solve it from a different angle, I get a wrong answer and can't understand where have I went wrong:



$$ a_n = S_{n+1} - S_{n} = k - frac{1}{3^{n+2}} - k +frac{1}{3^{n+1}}$$
$$ a_n = frac{1}{3^{n+1}} - frac{1}{3^{n+2}}$$
$$ a_n = frac{3^{n+2} - 3^{n+1}}{3^{n+1}cdot 3^{n+2}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n+3}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n}(3^3)} $$
$$ a_n = frac{3^{n - 2n}(3^2 - 3^1)}{3^3} = frac{3^{-n}(9 - 3)}{27} = frac{6}{27cdot 3^n} = frac{2}{9cdot 3^n} boxed{neq frac{2}{3^{n+2}}}$$



Help is greatly appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$




I am dealing with a general sequence, $a_n$, for which $S_n$ is the sequence of its partial sums.

I know that $a_n = S_{n} - S_{n-1}$, and therefore $a_n = S_{n+1} - S_{n}$
also.



I want to find $a_n$ given:
$$ S_n = k - frac{1}{3^{n+1}} $$



Here is the correct, confirmed calculation:
$$ a_n = S_{n} - S_{n-1} = k - frac{1}{3^{n+1}} - k +frac{1}{3^n}$$
$$ a_n = frac{1}{3^n} - frac{1}{3^{n+1}}$$
$$ a_n = frac{1-3^{-1}}{3^n} $$
$$ a_n = frac{2/3}{3^n} $$
$$ boxed{a_n = frac{2}{3^{n+1}}}$$



Fine and dandy, but when I try to solve it from a different angle, I get a wrong answer and can't understand where have I went wrong:



$$ a_n = S_{n+1} - S_{n} = k - frac{1}{3^{n+2}} - k +frac{1}{3^{n+1}}$$
$$ a_n = frac{1}{3^{n+1}} - frac{1}{3^{n+2}}$$
$$ a_n = frac{3^{n+2} - 3^{n+1}}{3^{n+1}cdot 3^{n+2}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n+3}} $$
$$ a_n = frac{3^n(3^2 - 3^1)}{3^{2n}(3^3)} $$
$$ a_n = frac{3^{n - 2n}(3^2 - 3^1)}{3^3} = frac{3^{-n}(9 - 3)}{27} = frac{6}{27cdot 3^n} = frac{2}{9cdot 3^n} boxed{neq frac{2}{3^{n+2}}}$$



Help is greatly appreciated.






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 17:11









OmG

2,512722




2,512722










asked Dec 22 '18 at 17:02









daedsidogdaedsidog

29017




29017








  • 3




    $begingroup$
    It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
    $endgroup$
    – Gerhard S.
    Dec 22 '18 at 17:05






  • 1




    $begingroup$
    Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
    $endgroup$
    – timtfj
    Dec 22 '18 at 17:20














  • 3




    $begingroup$
    It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
    $endgroup$
    – Gerhard S.
    Dec 22 '18 at 17:05






  • 1




    $begingroup$
    Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
    $endgroup$
    – timtfj
    Dec 22 '18 at 17:20








3




3




$begingroup$
It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
$endgroup$
– Gerhard S.
Dec 22 '18 at 17:05




$begingroup$
It is not true that $a_n=S_{n+1}-S_n$. What is true is $a_{n+1}=S_{n+1}-S_n$
$endgroup$
– Gerhard S.
Dec 22 '18 at 17:05




1




1




$begingroup$
Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
$endgroup$
– timtfj
Dec 22 '18 at 17:20




$begingroup$
Unsure why this had a downvote. It's a clear question with a clear answer, and a good example of how easy it can be to look in the wrong place for a mistake.
$endgroup$
– timtfj
Dec 22 '18 at 17:20










1 Answer
1






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oldest

votes


















1












$begingroup$

Your assumption is not correct in general! if $a_n = S_n - S_{n-1}$ then $a_{n+1} = S_{n+1} - S_n$. A contradiction for this is your example.



Your assumption will be true if the difference of each $S_{i} - S_{i-1}$ be a constant for all $i$. Hence, $S_{i+1} = S_i + c$ for all $i$. Therefore, $a_{n+1} = S_{n+1}-S_n = a_{n} =S_{n} - S_{n-1}= c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I am unsure what drove me to assume that in general.
    $endgroup$
    – daedsidog
    Dec 22 '18 at 17:14










  • $begingroup$
    @daedsidog my pleasure. take it easy. ; )
    $endgroup$
    – OmG
    Dec 22 '18 at 17:18











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1 Answer
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1 Answer
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active

oldest

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active

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active

oldest

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1












$begingroup$

Your assumption is not correct in general! if $a_n = S_n - S_{n-1}$ then $a_{n+1} = S_{n+1} - S_n$. A contradiction for this is your example.



Your assumption will be true if the difference of each $S_{i} - S_{i-1}$ be a constant for all $i$. Hence, $S_{i+1} = S_i + c$ for all $i$. Therefore, $a_{n+1} = S_{n+1}-S_n = a_{n} =S_{n} - S_{n-1}= c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I am unsure what drove me to assume that in general.
    $endgroup$
    – daedsidog
    Dec 22 '18 at 17:14










  • $begingroup$
    @daedsidog my pleasure. take it easy. ; )
    $endgroup$
    – OmG
    Dec 22 '18 at 17:18
















1












$begingroup$

Your assumption is not correct in general! if $a_n = S_n - S_{n-1}$ then $a_{n+1} = S_{n+1} - S_n$. A contradiction for this is your example.



Your assumption will be true if the difference of each $S_{i} - S_{i-1}$ be a constant for all $i$. Hence, $S_{i+1} = S_i + c$ for all $i$. Therefore, $a_{n+1} = S_{n+1}-S_n = a_{n} =S_{n} - S_{n-1}= c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I am unsure what drove me to assume that in general.
    $endgroup$
    – daedsidog
    Dec 22 '18 at 17:14










  • $begingroup$
    @daedsidog my pleasure. take it easy. ; )
    $endgroup$
    – OmG
    Dec 22 '18 at 17:18














1












1








1





$begingroup$

Your assumption is not correct in general! if $a_n = S_n - S_{n-1}$ then $a_{n+1} = S_{n+1} - S_n$. A contradiction for this is your example.



Your assumption will be true if the difference of each $S_{i} - S_{i-1}$ be a constant for all $i$. Hence, $S_{i+1} = S_i + c$ for all $i$. Therefore, $a_{n+1} = S_{n+1}-S_n = a_{n} =S_{n} - S_{n-1}= c$.






share|cite|improve this answer











$endgroup$



Your assumption is not correct in general! if $a_n = S_n - S_{n-1}$ then $a_{n+1} = S_{n+1} - S_n$. A contradiction for this is your example.



Your assumption will be true if the difference of each $S_{i} - S_{i-1}$ be a constant for all $i$. Hence, $S_{i+1} = S_i + c$ for all $i$. Therefore, $a_{n+1} = S_{n+1}-S_n = a_{n} =S_{n} - S_{n-1}= c$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 17:10

























answered Dec 22 '18 at 17:04









OmGOmG

2,512722




2,512722












  • $begingroup$
    Thank you. I am unsure what drove me to assume that in general.
    $endgroup$
    – daedsidog
    Dec 22 '18 at 17:14










  • $begingroup$
    @daedsidog my pleasure. take it easy. ; )
    $endgroup$
    – OmG
    Dec 22 '18 at 17:18


















  • $begingroup$
    Thank you. I am unsure what drove me to assume that in general.
    $endgroup$
    – daedsidog
    Dec 22 '18 at 17:14










  • $begingroup$
    @daedsidog my pleasure. take it easy. ; )
    $endgroup$
    – OmG
    Dec 22 '18 at 17:18
















$begingroup$
Thank you. I am unsure what drove me to assume that in general.
$endgroup$
– daedsidog
Dec 22 '18 at 17:14




$begingroup$
Thank you. I am unsure what drove me to assume that in general.
$endgroup$
– daedsidog
Dec 22 '18 at 17:14












$begingroup$
@daedsidog my pleasure. take it easy. ; )
$endgroup$
– OmG
Dec 22 '18 at 17:18




$begingroup$
@daedsidog my pleasure. take it easy. ; )
$endgroup$
– OmG
Dec 22 '18 at 17:18


















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