Convert this sum to normal expression
$begingroup$
If I have a sum like this:
$$2sum_{i=0}^{n-1}3^i(3^{n-i}-1)$$ How do I convert it so that I can lose the sum. For example if it was
$$sum_{i=0}^{n}n$$
then the result would be
$$frac{n(n-1)}{2}$$
Is there a general principle how to do this, for example like for geometric sums?
discrete-mathematics summation systems-of-equations
edited Dec 17 '18 at 18:57
pwerth
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
If I have a sum like this:
$$2sum_{i=0}^{n-1}3^i(3^{n-i}-1)$$ How do I convert it so that I can lose the sum. For example if it was
$$sum_{i=0}^{n}n$$
then the result would be
$$frac{n(n-1)}{2}$$
Is there a general principle how to do this, for example like for geometric sums?
discrete-mathematics summation systems-of-equations
edited Dec 17 '18 at 18:57
pwerth
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
$endgroup$
2
$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03
add a comment |
$begingroup$
If I have a sum like this:
$$2sum_{i=0}^{n-1}3^i(3^{n-i}-1)$$ How do I convert it so that I can lose the sum. For example if it was
$$sum_{i=0}^{n}n$$
then the result would be
$$frac{n(n-1)}{2}$$
Is there a general principle how to do this, for example like for geometric sums?
discrete-mathematics summation systems-of-equations
edited Dec 17 '18 at 18:57
pwerth
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
$endgroup$
If I have a sum like this:
$$2sum_{i=0}^{n-1}3^i(3^{n-i}-1)$$ How do I convert it so that I can lose the sum. For example if it was
$$sum_{i=0}^{n}n$$
then the result would be
$$frac{n(n-1)}{2}$$
Is there a general principle how to do this, for example like for geometric sums?
discrete-mathematics summation systems-of-equations
discrete-mathematics summation systems-of-equations
edited Dec 17 '18 at 18:57
pwerth
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
edited Dec 17 '18 at 18:57
pwerth
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
edited Dec 17 '18 at 18:57
pwerth
3,300417
edited Dec 17 '18 at 18:57
pwerth
3,300417
edited Dec 17 '18 at 18:57
pwerth
3,300417
3,300417
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
asked Dec 17 '18 at 18:42
ponikoliponikoli
416
416
2
$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03
add a comment |
2
$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03
2
2
$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.
We obtain
begin{align*}
color{blue}{2sum_{i=0}^{n-1}3^ileft(3^{n-i}-1right)}&=
2sum_{i=0}^{n-1}left(3^n-3^iright)\
&=2cdot3^nsum_{i=0}^{n-1}1-2sum_{i=0}^{n-1}3^i\
&=2cdot 3^nn-2cdotfrac{3^n-1}{3-1}\
&=2cdot 3^nn-3^n+1\
&,,color{blue}{=3^n(2n-1)+1}
end{align*}
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.
We obtain
begin{align*}
color{blue}{2sum_{i=0}^{n-1}3^ileft(3^{n-i}-1right)}&=
2sum_{i=0}^{n-1}left(3^n-3^iright)\
&=2cdot3^nsum_{i=0}^{n-1}1-2sum_{i=0}^{n-1}3^i\
&=2cdot 3^nn-2cdotfrac{3^n-1}{3-1}\
&=2cdot 3^nn-3^n+1\
&,,color{blue}{=3^n(2n-1)+1}
end{align*}
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
$endgroup$
add a comment |
$begingroup$
The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.
We obtain
begin{align*}
color{blue}{2sum_{i=0}^{n-1}3^ileft(3^{n-i}-1right)}&=
2sum_{i=0}^{n-1}left(3^n-3^iright)\
&=2cdot3^nsum_{i=0}^{n-1}1-2sum_{i=0}^{n-1}3^i\
&=2cdot 3^nn-2cdotfrac{3^n-1}{3-1}\
&=2cdot 3^nn-3^n+1\
&,,color{blue}{=3^n(2n-1)+1}
end{align*}
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
$endgroup$
add a comment |
$begingroup$
The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.
We obtain
begin{align*}
color{blue}{2sum_{i=0}^{n-1}3^ileft(3^{n-i}-1right)}&=
2sum_{i=0}^{n-1}left(3^n-3^iright)\
&=2cdot3^nsum_{i=0}^{n-1}1-2sum_{i=0}^{n-1}3^i\
&=2cdot 3^nn-2cdotfrac{3^n-1}{3-1}\
&=2cdot 3^nn-3^n+1\
&,,color{blue}{=3^n(2n-1)+1}
end{align*}
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
$endgroup$
The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.
We obtain
begin{align*}
color{blue}{2sum_{i=0}^{n-1}3^ileft(3^{n-i}-1right)}&=
2sum_{i=0}^{n-1}left(3^n-3^iright)\
&=2cdot3^nsum_{i=0}^{n-1}1-2sum_{i=0}^{n-1}3^i\
&=2cdot 3^nn-2cdotfrac{3^n-1}{3-1}\
&=2cdot 3^nn-3^n+1\
&,,color{blue}{=3^n(2n-1)+1}
end{align*}
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
answered Dec 17 '18 at 20:55
Markus ScheuerMarkus Scheuer
62.8k460150
62.8k460150
add a comment |
add a comment |
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$begingroup$
You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please.
$endgroup$
– drhab
Dec 17 '18 at 18:53
$begingroup$
Also note that $sum_{i=0}^n n = n + cdots + n = (n+1)n$ and $sum_{i=0}^n i = 0+1+cdots+n = frac{n(color{red}{n+1})}{2}$.
$endgroup$
– Christoph
Dec 17 '18 at 21:03