G.P. = [itex] \sum_{n=1}^k a_{1}(\frac{a_{n+1}}{a_{n}})^{n-1} [/itex]

[Lebombo]

A geometric series is $S_{n}= a_{1} + a_{2} + a_{3} + ...+ a_{n} = a_{1}r^{0} + a_{2}r^{1} + a_{3}r^{2} + ...+ a_{n}r^{n-1}$

In a geometric series, the first term = $a_{1}$

The common ratio = $\frac{a_{n+1}}{a_{n}}$

The nth term of a geometric sequence is $a_{n}= a_{1}r^{n-1}= a_{1}(\frac{a_{n+1}}{a_{n}})^{n-1}$

The question is:

In the summand of the sigma notation, can the general term of geometric sequence formula be expressed this way:

$$S_{n}= \sum_{i=1}^n a_{1}r^{i-1} = \sum_{i=1}^n a_{1}(\frac{a_{i+1}}{a_{i}})^{i-1}$$

 

[haruspex]

It can, but why would you want to? It loses the information that the ratio is a constant.

 

[Mark44]

A geometric series is $S_{n}= a_{1} + a_{2} + a_{3} + ...+ a_{n} = a_{1}r^{0} + a_{2}r^{1} + a_{3}r^{2} + ...+ a_{n}r^{n-1}$

The two expressions you show aren't equal unless r = 1.

You should write your (finite) geometric series like so:
$S_{n}= ar^{0} + ar^{1} + ar^{2} + ...+ ar^{n-1}$

In a geometric series, the first term = $a_{1}$

There is no need for subscripts in a geometric series. The first term is a, the next is ar, the one after that is ar², and so on.

The common ratio = $\frac{a_{n+1}}{a_{n}}$

The common ratio is r. Each term in the series is r times the previous term.

The nth term of a geometric sequence is $a_{n}= a_{1}r^{n-1}= a_{1}(\frac{a_{n+1}}{a_{n}})^{n-1}$

The n-th term is ar^n-1.

The question is:

In the summand of the sigma notation, can the general term of geometric sequence formula be expressed this way:

$$S_{n}= \sum_{i=1}^n a_{1}r^{i-1} = \sum_{i=1}^n a_{1}(\frac{a_{i+1}}{a_{i}})^{i-1}$$

It's much simpler like this:
$$S_n = \sum_{k = 0}^{n-1} ar^k$$