Mixing
Calculating sum of consecutive powers of a number
$begingroup$
Here is my problem, I want to compute the $$sum_{i=0}^n P^i : Pin ℤ_{>1}$$
I know I can implement it using an easy recursive function, but since I want to use the formula in a spreadsheet, is there a better approach to this?
Thanks.
summation exponentiation
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
$endgroup$
add a comment |
$begingroup$
Here is my problem, I want to compute the $$sum_{i=0}^n P^i : Pin ℤ_{>1}$$
I know I can implement it using an easy recursive function, but since I want to use the formula in a spreadsheet, is there a better approach to this?
Thanks.
summation exponentiation
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
$endgroup$
add a comment |
$begingroup$
Here is my problem, I want to compute the $$sum_{i=0}^n P^i : Pin ℤ_{>1}$$
I know I can implement it using an easy recursive function, but since I want to use the formula in a spreadsheet, is there a better approach to this?
Thanks.
summation exponentiation
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
$endgroup$
Here is my problem, I want to compute the $$sum_{i=0}^n P^i : Pin ℤ_{>1}$$
I know I can implement it using an easy recursive function, but since I want to use the formula in a spreadsheet, is there a better approach to this?
Thanks.
summation exponentiation
summation exponentiation
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
asked Oct 13 '14 at 14:44
sasanjsasanj
58114
58114
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If we call the sum $S_n$, then $$P cdot S_n = P + P^2 + P^3 + cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$
Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = frac{P^{n+1}-1}{P-1}$$
This is a partial sum of a geometric series.
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
$endgroup$
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
add a comment |
$begingroup$
The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :
$$sum_{i=0}^n P^i = dfrac{P^{n+1} -1}{P-1}$$
EDIT : Let's prove this !
$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$
You have the result by dividing both sides by $P-1$.
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
$endgroup$
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
add a comment |
$begingroup$
That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $displaystyle frac{P^{n+1}-1}{P-1}$
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
We have
$$begin{array}{l}
S_n&=1+P+P^2+P^3+cdots+P^n\
Pcdot S_n&=0+P+P^2+P^3+cdots+P^n+P^{n+1}
end{array}$$
Subtracting two above equations gives
$$
S_n-Pcdot S_n=1-P^{n+1}
$$
divide by $S_n$
$$
1-P=dfrac{1-P^{n+1}}{S_n}\
S_n=dfrac{1-P^{n+1}}{1-P}
$$
answered Oct 21 '17 at 13:39
BrightBright
200111
$endgroup$
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we call the sum $S_n$, then $$P cdot S_n = P + P^2 + P^3 + cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$
Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = frac{P^{n+1}-1}{P-1}$$
This is a partial sum of a geometric series.
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
$endgroup$
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
add a comment |
$begingroup$
If we call the sum $S_n$, then $$P cdot S_n = P + P^2 + P^3 + cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$
Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = frac{P^{n+1}-1}{P-1}$$
This is a partial sum of a geometric series.
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
$endgroup$
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
add a comment |
$begingroup$
If we call the sum $S_n$, then $$P cdot S_n = P + P^2 + P^3 + cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$
Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = frac{P^{n+1}-1}{P-1}$$
This is a partial sum of a geometric series.
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
$endgroup$
If we call the sum $S_n$, then $$P cdot S_n = P + P^2 + P^3 + cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$
Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = frac{P^{n+1}-1}{P-1}$$
This is a partial sum of a geometric series.
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
answered Oct 13 '14 at 14:50
JoelJoel
14.3k12240
14.3k12240
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
add a comment |
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
3
3
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
Thank you very much. What a dump I was! Back to elementary mathematics.
$endgroup$
– sasanj
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
$begingroup$
I am glad I could help
$endgroup$
– Joel
Oct 13 '14 at 14:56
add a comment |
$begingroup$
The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :
$$sum_{i=0}^n P^i = dfrac{P^{n+1} -1}{P-1}$$
EDIT : Let's prove this !
$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$
You have the result by dividing both sides by $P-1$.
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
$endgroup$
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
add a comment |
$begingroup$
The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :
$$sum_{i=0}^n P^i = dfrac{P^{n+1} -1}{P-1}$$
EDIT : Let's prove this !
$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$
You have the result by dividing both sides by $P-1$.
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
$endgroup$
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
add a comment |
$begingroup$
The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :
$$sum_{i=0}^n P^i = dfrac{P^{n+1} -1}{P-1}$$
EDIT : Let's prove this !
$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$
You have the result by dividing both sides by $P-1$.
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
$endgroup$
The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :
$$sum_{i=0}^n P^i = dfrac{P^{n+1} -1}{P-1}$$
EDIT : Let's prove this !
$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$
You have the result by dividing both sides by $P-1$.
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
edited Oct 13 '14 at 14:55
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
answered Oct 13 '14 at 14:48
TraklonTraklon
2,70711127
2,70711127
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
add a comment |
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
$begingroup$
Thanks for answering that fast. But is there a mathematical proof you can point me to?
$endgroup$
– sasanj
Oct 13 '14 at 14:50
add a comment |
$begingroup$
That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $displaystyle frac{P^{n+1}-1}{P-1}$
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $displaystyle frac{P^{n+1}-1}{P-1}$
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $displaystyle frac{P^{n+1}-1}{P-1}$
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
$endgroup$
That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $displaystyle frac{P^{n+1}-1}{P-1}$
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
answered Oct 13 '14 at 14:50
DeepakDeepak
16.8k11436
16.8k11436
add a comment |
add a comment |
$begingroup$
We have
$$begin{array}{l}
S_n&=1+P+P^2+P^3+cdots+P^n\
Pcdot S_n&=0+P+P^2+P^3+cdots+P^n+P^{n+1}
end{array}$$
Subtracting two above equations gives
$$
S_n-Pcdot S_n=1-P^{n+1}
$$
divide by $S_n$
$$
1-P=dfrac{1-P^{n+1}}{S_n}\
S_n=dfrac{1-P^{n+1}}{1-P}
$$
answered Oct 21 '17 at 13:39
BrightBright
200111
$endgroup$
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
add a comment |
$begingroup$
We have
$$begin{array}{l}
S_n&=1+P+P^2+P^3+cdots+P^n\
Pcdot S_n&=0+P+P^2+P^3+cdots+P^n+P^{n+1}
end{array}$$
Subtracting two above equations gives
$$
S_n-Pcdot S_n=1-P^{n+1}
$$
divide by $S_n$
$$
1-P=dfrac{1-P^{n+1}}{S_n}\
S_n=dfrac{1-P^{n+1}}{1-P}
$$
answered Oct 21 '17 at 13:39
BrightBright
200111
$endgroup$
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
add a comment |
$begingroup$
We have
$$begin{array}{l}
S_n&=1+P+P^2+P^3+cdots+P^n\
Pcdot S_n&=0+P+P^2+P^3+cdots+P^n+P^{n+1}
end{array}$$
Subtracting two above equations gives
$$
S_n-Pcdot S_n=1-P^{n+1}
$$
divide by $S_n$
$$
1-P=dfrac{1-P^{n+1}}{S_n}\
S_n=dfrac{1-P^{n+1}}{1-P}
$$
answered Oct 21 '17 at 13:39
BrightBright
200111
$endgroup$
We have
$$begin{array}{l}
S_n&=1+P+P^2+P^3+cdots+P^n\
Pcdot S_n&=0+P+P^2+P^3+cdots+P^n+P^{n+1}
end{array}$$
Subtracting two above equations gives
$$
S_n-Pcdot S_n=1-P^{n+1}
$$
divide by $S_n$
$$
1-P=dfrac{1-P^{n+1}}{S_n}\
S_n=dfrac{1-P^{n+1}}{1-P}
$$
answered Oct 21 '17 at 13:39
BrightBright
200111
edited Oct 21 '17 at 13:46
answered Oct 21 '17 at 13:39
BrightBright
200111
answered Oct 21 '17 at 13:39
BrightBright
200111
answered Oct 21 '17 at 13:39
BrightBright
200111
200111
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
add a comment |
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
$begingroup$
This is of course correct. But it's a duplicate of the three other good answers to this three year old question. You're an experienced user of the site, asking questions and accepting answers. Why not spend your answering time on new questions?
$endgroup$
– Ethan Bolker
Oct 21 '17 at 13:50
add a comment |
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