Can this be written in sigma notation: a + ar + ar^2 + + ar^n-1?

[Lebombo]

Homework Statement

$$a_{1}r^{1} + a_{1}r^{2} + a_{1}r^{3} +... + a_{1}r^{n-3} + a_{1}r^{n-2} + a_{1}r^{n-1} = \sum_{n=1}^{?} a_{1}r^{n-1}$$

What value would replace the "?"


2. The attempt at a solution

EDIT: I edited this post after receiving your reply Cepheid.

My gut would say it is possible, but in thinking about it, it seems there is no way to represent it.

Guess 1)
Perhaps it's not possible to write the general terms of geometric sequences in sigma notation.

If it's not possible to write this in sigma notation, is it because the general term of geometric sequence formula, $a_{1}r^{n-1}$, represents a "term" and not a "function?"

Guess 2)
It is possible to put the general term of geometric sequence formula, $a_{1}r^{n-1}$, into sigma notation. However instead of representing: $$a_{1}r^{1} + a_{1}r^{2} + a_{1}r^{3} +... + a_{1}r^{n-3} + a_{1}r^{n-2} + a_{1}r^{n-1} = \sum_{n=1}^{N} a_{1}r^{n-1}$$

It will instead represent this: $\sum_{n=1}^{N} a_{1}r^{n-1} = a_{1}r^{n-1} + a_{1}r^{n-1} + a_{1}r^{n-1} + ... + a_{1}r^{n-1}$ Where $a_{1}r^{n-1}$ is simply repeated N times. Similar to $\sum_{n=1}^{N} 5$ = 5 + 5 + 5 +...+ 5.

 

[cepheid]

Am I missing something? $$a_1 \sum_{i=1}^{n-1} r^i$$

 

[Lebombo]

So the summand, $a_1 \sum_{i=1}^{n-1} (summand)$, does not include i-1 in the exponent, but rather the exponent i, while n-1 is written in the upper bound.

Makes sense, so would that mean if the general term of the geometric sequence formula were to be put in the summand, while written in the upper bound was an arbitrary N, as in $\sum_{i=1}^{N} a_{n}r^{n-1}$ , then the outcome would not be a geometric series, but rather a repeating series, where $a_{1}r^{n-1}$ simply repeats just as the number 5 in the summand would simply repeat N times?

And suppose I want to add up the first 10 terms of something like $2^{n}$.

This is written $\sum_{k=1}^{10} 2^{n} = 2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{10}= 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024$

Since the nth term of a geometric sequence is $$a_{n}=a_{1}r^{n-1} = a_1\sum_{i=1}^{n-1} r^i = \sum_{i=1}^{n-1} (2)2^i$$

Does this mean that if I want to add the first 10 terms, where n=10, then the sigma notation would be: $\sum_{i=1}^{10-1} (2)2^i = (2)2^1 + (2)2^2 + (2)2^3 +...+(2)2^9 = 4 + 8+ 16 + ...+ 1024$.

Since the first term in this sequence is 4 rather than 2, perhaps the sigma notation should include a i=0 in the index instead of i=1 to make it 10 total terms in the summation. That is: $$a_1 \sum_{i=0}^{n-1} r^i$$

 

[Ray Vickson]

Homework Statement

$$a_{1}r^{1} + a_{1}r^{2} + a_{1}r^{3} +... + a_{1}r^{n-3} + a_{1}r^{n-2} + a_{1}r^{n-1} = \sum_{n=1}^{?} a_{1}r^{n-1}$$

What value would replace the "?"


2. The attempt at a solution

EDIT: I edited this post after receiving your reply Cepheid.

My gut would say it is possible, but in thinking about it, it seems there is no way to represent it.

Guess 1)
Perhaps it's not possible to write the general terms of geometric sequences in sigma notation.

If it's not possible to write this in sigma notation, is it because the general term of geometric sequence formula, $a_{1}r^{n-1}$, represents a "term" and not a "function?"

Guess 2)
It is possible to put the general term of geometric sequence formula, $a_{1}r^{n-1}$, into sigma notation. However instead of representing: $$a_{1}r^{1} + a_{1}r^{2} + a_{1}r^{3} +... + a_{1}r^{n-3} + a_{1}r^{n-2} + a_{1}r^{n-1} = \sum_{n=1}^{N} a_{1}r^{n-1}$$

It will instead represent this: $\sum_{n=1}^{N} a_{1}r^{n-1} = a_{1}r^{n-1} + a_{1}r^{n-1} + a_{1}r^{n-1} + ... + a_{1}r^{n-1}$ Where $a_{1}r^{n-1}$ is simply repeated N times. Similar to $\sum_{n=1}^{N} 5$ = 5 + 5 + 5 +...+ 5.

The summation index is any "dummy symbol" except 'n'; you cannot use 'n' because that is already reserved to specify the last term $r^{n-1}$. So, if $S_n$ is your sum, we can write
$$S_n = \sum_{i=1}^{n-1} a_1 r^i = \sum_{j=1}^{n-1} a_1 r^j = \cdots =\sum_{\text{anything}=1}^{\text{anything} = n-1} a_1 r^{\,\text{anything}}.$$