How do I convert a sum (from $1$ to $n$) of powers into a function of $n$?

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This question is going to need some context. The TL;DR is:



How do I get a function equal to $sum_i=1^n1.15^i$ without using sums?




I'm using Google sheets to keep track of something and as far as I know it doesn't support sequences or sums in one line (I don't want to have every terms of sequence in one cell and then sum the cells). Though this isn't a question about how to do things in Google sheets. Last time I had a sequence I managed to find a function using simple factors with Wolfram Alpha, equal to the sequence.



The first sequence is :




$A_n = A_n-1times1.15$



$A_0 = a $




$A_0$ is variable and goes in a cell. For this sequence, I use $A_n = A_0times1.15^n$, or as a function $f_a(n)=atimes1.15^n$.



The second sequence $B$ is the sum of all the terms of $A$.




$B_n = sum_i=1^n A_i $



$ B_n = sum_i=1^n A_0times1.15^i$



$ B_n = A_0 sum_i=1^n 1.15^i$




And that is problematic to calculate with a "google sheets friendly" function. For example, $B_3$ goes like this:




$B_3 = A_3 + A_2 + A_1 + A_0$



$B_3 = atimes1.15^3 + atimes1.15^2 + atimes1.15 + a$



$B_3 = a (1.15^3 + 1.15^2 + 1.15 + 1)$




So I have get this series of polynonial-looking terms and I have no idea how I can write a function that calculates $B_3$ and $B_n$ easily, without having to use a cell for each term. I've reduced my problem to calculating the sum I gave in the TL;DR but I'm stuck there.



If it's unclear what I'm asking, I just want to know if there's an easy way to calculate $sum_i=1^n1.15^i$ for a given $n$ without using sums (i.e. without having to calculate each term of the sum). "No" is an acceptable answer, and any other way to express the same thing is welcome.



Also, is there a name for the act of finding an expression of $U_n$ that uses only $U_0$ instead of $U_n-1$?







share|cite|improve this question


















  • 5




    This is a standard geometric series and closed formulas are known (see the link).
    – lulu
    Aug 27 at 0:45






  • 1




    @lulu post the geometric series sum as an answer for the newbie who's not studying math?
    – Ethan Bolker
    Aug 27 at 0:57










  • @lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
    – Teleporting Goat
    Aug 27 at 0:59










  • Who uses comma to denote the decimal separator?
    – tst
    Aug 27 at 1:52










  • @tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
    – Teleporting Goat
    Aug 27 at 2:03














up vote
1
down vote

favorite












This question is going to need some context. The TL;DR is:



How do I get a function equal to $sum_i=1^n1.15^i$ without using sums?




I'm using Google sheets to keep track of something and as far as I know it doesn't support sequences or sums in one line (I don't want to have every terms of sequence in one cell and then sum the cells). Though this isn't a question about how to do things in Google sheets. Last time I had a sequence I managed to find a function using simple factors with Wolfram Alpha, equal to the sequence.



The first sequence is :




$A_n = A_n-1times1.15$



$A_0 = a $




$A_0$ is variable and goes in a cell. For this sequence, I use $A_n = A_0times1.15^n$, or as a function $f_a(n)=atimes1.15^n$.



The second sequence $B$ is the sum of all the terms of $A$.




$B_n = sum_i=1^n A_i $



$ B_n = sum_i=1^n A_0times1.15^i$



$ B_n = A_0 sum_i=1^n 1.15^i$




And that is problematic to calculate with a "google sheets friendly" function. For example, $B_3$ goes like this:




$B_3 = A_3 + A_2 + A_1 + A_0$



$B_3 = atimes1.15^3 + atimes1.15^2 + atimes1.15 + a$



$B_3 = a (1.15^3 + 1.15^2 + 1.15 + 1)$




So I have get this series of polynonial-looking terms and I have no idea how I can write a function that calculates $B_3$ and $B_n$ easily, without having to use a cell for each term. I've reduced my problem to calculating the sum I gave in the TL;DR but I'm stuck there.



If it's unclear what I'm asking, I just want to know if there's an easy way to calculate $sum_i=1^n1.15^i$ for a given $n$ without using sums (i.e. without having to calculate each term of the sum). "No" is an acceptable answer, and any other way to express the same thing is welcome.



Also, is there a name for the act of finding an expression of $U_n$ that uses only $U_0$ instead of $U_n-1$?







share|cite|improve this question


















  • 5




    This is a standard geometric series and closed formulas are known (see the link).
    – lulu
    Aug 27 at 0:45






  • 1




    @lulu post the geometric series sum as an answer for the newbie who's not studying math?
    – Ethan Bolker
    Aug 27 at 0:57










  • @lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
    – Teleporting Goat
    Aug 27 at 0:59










  • Who uses comma to denote the decimal separator?
    – tst
    Aug 27 at 1:52










  • @tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
    – Teleporting Goat
    Aug 27 at 2:03












up vote
1
down vote

favorite









up vote
1
down vote

favorite











This question is going to need some context. The TL;DR is:



How do I get a function equal to $sum_i=1^n1.15^i$ without using sums?




I'm using Google sheets to keep track of something and as far as I know it doesn't support sequences or sums in one line (I don't want to have every terms of sequence in one cell and then sum the cells). Though this isn't a question about how to do things in Google sheets. Last time I had a sequence I managed to find a function using simple factors with Wolfram Alpha, equal to the sequence.



The first sequence is :




$A_n = A_n-1times1.15$



$A_0 = a $




$A_0$ is variable and goes in a cell. For this sequence, I use $A_n = A_0times1.15^n$, or as a function $f_a(n)=atimes1.15^n$.



The second sequence $B$ is the sum of all the terms of $A$.




$B_n = sum_i=1^n A_i $



$ B_n = sum_i=1^n A_0times1.15^i$



$ B_n = A_0 sum_i=1^n 1.15^i$




And that is problematic to calculate with a "google sheets friendly" function. For example, $B_3$ goes like this:




$B_3 = A_3 + A_2 + A_1 + A_0$



$B_3 = atimes1.15^3 + atimes1.15^2 + atimes1.15 + a$



$B_3 = a (1.15^3 + 1.15^2 + 1.15 + 1)$




So I have get this series of polynonial-looking terms and I have no idea how I can write a function that calculates $B_3$ and $B_n$ easily, without having to use a cell for each term. I've reduced my problem to calculating the sum I gave in the TL;DR but I'm stuck there.



If it's unclear what I'm asking, I just want to know if there's an easy way to calculate $sum_i=1^n1.15^i$ for a given $n$ without using sums (i.e. without having to calculate each term of the sum). "No" is an acceptable answer, and any other way to express the same thing is welcome.



Also, is there a name for the act of finding an expression of $U_n$ that uses only $U_0$ instead of $U_n-1$?







share|cite|improve this question














This question is going to need some context. The TL;DR is:



How do I get a function equal to $sum_i=1^n1.15^i$ without using sums?




I'm using Google sheets to keep track of something and as far as I know it doesn't support sequences or sums in one line (I don't want to have every terms of sequence in one cell and then sum the cells). Though this isn't a question about how to do things in Google sheets. Last time I had a sequence I managed to find a function using simple factors with Wolfram Alpha, equal to the sequence.



The first sequence is :




$A_n = A_n-1times1.15$



$A_0 = a $




$A_0$ is variable and goes in a cell. For this sequence, I use $A_n = A_0times1.15^n$, or as a function $f_a(n)=atimes1.15^n$.



The second sequence $B$ is the sum of all the terms of $A$.




$B_n = sum_i=1^n A_i $



$ B_n = sum_i=1^n A_0times1.15^i$



$ B_n = A_0 sum_i=1^n 1.15^i$




And that is problematic to calculate with a "google sheets friendly" function. For example, $B_3$ goes like this:




$B_3 = A_3 + A_2 + A_1 + A_0$



$B_3 = atimes1.15^3 + atimes1.15^2 + atimes1.15 + a$



$B_3 = a (1.15^3 + 1.15^2 + 1.15 + 1)$




So I have get this series of polynonial-looking terms and I have no idea how I can write a function that calculates $B_3$ and $B_n$ easily, without having to use a cell for each term. I've reduced my problem to calculating the sum I gave in the TL;DR but I'm stuck there.



If it's unclear what I'm asking, I just want to know if there's an easy way to calculate $sum_i=1^n1.15^i$ for a given $n$ without using sums (i.e. without having to calculate each term of the sum). "No" is an acceptable answer, and any other way to express the same thing is welcome.



Also, is there a name for the act of finding an expression of $U_n$ that uses only $U_0$ instead of $U_n-1$?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 8:32

























asked Aug 27 at 0:39









Teleporting Goat

1085




1085







  • 5




    This is a standard geometric series and closed formulas are known (see the link).
    – lulu
    Aug 27 at 0:45






  • 1




    @lulu post the geometric series sum as an answer for the newbie who's not studying math?
    – Ethan Bolker
    Aug 27 at 0:57










  • @lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
    – Teleporting Goat
    Aug 27 at 0:59










  • Who uses comma to denote the decimal separator?
    – tst
    Aug 27 at 1:52










  • @tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
    – Teleporting Goat
    Aug 27 at 2:03












  • 5




    This is a standard geometric series and closed formulas are known (see the link).
    – lulu
    Aug 27 at 0:45






  • 1




    @lulu post the geometric series sum as an answer for the newbie who's not studying math?
    – Ethan Bolker
    Aug 27 at 0:57










  • @lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
    – Teleporting Goat
    Aug 27 at 0:59










  • Who uses comma to denote the decimal separator?
    – tst
    Aug 27 at 1:52










  • @tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
    – Teleporting Goat
    Aug 27 at 2:03







5




5




This is a standard geometric series and closed formulas are known (see the link).
– lulu
Aug 27 at 0:45




This is a standard geometric series and closed formulas are known (see the link).
– lulu
Aug 27 at 0:45




1




1




@lulu post the geometric series sum as an answer for the newbie who's not studying math?
– Ethan Bolker
Aug 27 at 0:57




@lulu post the geometric series sum as an answer for the newbie who's not studying math?
– Ethan Bolker
Aug 27 at 0:57












@lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
– Teleporting Goat
Aug 27 at 0:59




@lulu Thanks, but I can't seem to find what I'm looking for. There's only a small bit about Geometric power series, and I don't want a formula to infinity, I want something that works for a specific $n$.
– Teleporting Goat
Aug 27 at 0:59












Who uses comma to denote the decimal separator?
– tst
Aug 27 at 1:52




Who uses comma to denote the decimal separator?
– tst
Aug 27 at 1:52












@tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
– Teleporting Goat
Aug 27 at 2:03




@tst A number of languages, like French or German for example (see docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html). But this site is in English I should have used a dot.
– Teleporting Goat
Aug 27 at 2:03










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










The formula you want is $$sum_i=0^n x^i = frac1-x^n+11-x.$$



This can be seen as follows:



beginalign*
(1-x)(1 + x + cdots + x^n) &= (1 + x + dots + x^n) - x(1 + x + dots + x^n)\
&=(1 + x + dots + x^n) - (x + x^2 + dots + x^n+1)\
&= 1 - x^n+1.
endalign*



Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).






share|cite|improve this answer




















  • Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
    – Teleporting Goat
    Aug 27 at 1:09










  • I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
    – Jair Taylor
    Aug 27 at 1:10










  • Thanks, "partial sum" is what I was looking for :)
    – Teleporting Goat
    Aug 27 at 1:20










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










The formula you want is $$sum_i=0^n x^i = frac1-x^n+11-x.$$



This can be seen as follows:



beginalign*
(1-x)(1 + x + cdots + x^n) &= (1 + x + dots + x^n) - x(1 + x + dots + x^n)\
&=(1 + x + dots + x^n) - (x + x^2 + dots + x^n+1)\
&= 1 - x^n+1.
endalign*



Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).






share|cite|improve this answer




















  • Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
    – Teleporting Goat
    Aug 27 at 1:09










  • I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
    – Jair Taylor
    Aug 27 at 1:10










  • Thanks, "partial sum" is what I was looking for :)
    – Teleporting Goat
    Aug 27 at 1:20














up vote
2
down vote



accepted










The formula you want is $$sum_i=0^n x^i = frac1-x^n+11-x.$$



This can be seen as follows:



beginalign*
(1-x)(1 + x + cdots + x^n) &= (1 + x + dots + x^n) - x(1 + x + dots + x^n)\
&=(1 + x + dots + x^n) - (x + x^2 + dots + x^n+1)\
&= 1 - x^n+1.
endalign*



Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).






share|cite|improve this answer




















  • Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
    – Teleporting Goat
    Aug 27 at 1:09










  • I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
    – Jair Taylor
    Aug 27 at 1:10










  • Thanks, "partial sum" is what I was looking for :)
    – Teleporting Goat
    Aug 27 at 1:20












up vote
2
down vote



accepted







up vote
2
down vote



accepted






The formula you want is $$sum_i=0^n x^i = frac1-x^n+11-x.$$



This can be seen as follows:



beginalign*
(1-x)(1 + x + cdots + x^n) &= (1 + x + dots + x^n) - x(1 + x + dots + x^n)\
&=(1 + x + dots + x^n) - (x + x^2 + dots + x^n+1)\
&= 1 - x^n+1.
endalign*



Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).






share|cite|improve this answer












The formula you want is $$sum_i=0^n x^i = frac1-x^n+11-x.$$



This can be seen as follows:



beginalign*
(1-x)(1 + x + cdots + x^n) &= (1 + x + dots + x^n) - x(1 + x + dots + x^n)\
&=(1 + x + dots + x^n) - (x + x^2 + dots + x^n+1)\
&= 1 - x^n+1.
endalign*



Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 27 at 1:05









Jair Taylor

8,50932144




8,50932144











  • Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
    – Teleporting Goat
    Aug 27 at 1:09










  • I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
    – Jair Taylor
    Aug 27 at 1:10










  • Thanks, "partial sum" is what I was looking for :)
    – Teleporting Goat
    Aug 27 at 1:20
















  • Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
    – Teleporting Goat
    Aug 27 at 1:09










  • I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
    – Jair Taylor
    Aug 27 at 1:10










  • Thanks, "partial sum" is what I was looking for :)
    – Teleporting Goat
    Aug 27 at 1:20















Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
– Teleporting Goat
Aug 27 at 1:09




Thanks, this is the formula I want! Does it have a name? (The form on the right I mean)
– Teleporting Goat
Aug 27 at 1:09












I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
– Jair Taylor
Aug 27 at 1:10




I don't know if there's a name for this formula. You could call the partial sum of a geometric series, a geometric sum, or somesuch.
– Jair Taylor
Aug 27 at 1:10












Thanks, "partial sum" is what I was looking for :)
– Teleporting Goat
Aug 27 at 1:20




Thanks, "partial sum" is what I was looking for :)
– Teleporting Goat
Aug 27 at 1:20

















 

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