Understanding the Geometric Series Sum

The question asks us to find the sum of the finite geometric series represented by the notation $\sum_{n=0}^{r} q^n$. This series expands to $1 + q + q^2 + \dots + q^r$. We need to determine the correct formula for this sum.

Recalling the Finite Geometric Series Formula

A finite geometric series is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The standard formula for the sum ($S_N$) of the first $N$ terms of a geometric series is:

$ S_N = \frac{a(1-r^N)}{1-r} $

where:

• $a$ is the first term
• $r$ is the common ratio
• $N$ is the number of terms

This formula is valid when the common ratio $r \neq 1$.

Applying the Formula to the Given Series

Let's identify the components of our specific series, $\sum_{n=0}^{r} q^n = 1 + q + q^2 + \dots + q^r$:

• The first term ($a$) is $q^0 = 1$.
• The common ratio is $q$.
• The number of terms ($N$) is determined by the index range. Since $n$ goes from $0$ to $r$, there are $(r - 0) + 1 = r+1$ terms.

Now, we substitute these values into the standard formula:

$ S = \frac{1 \cdot (1 - q^{(r+1)})}{1 - q} $

$ S = \frac{1 - q^{r+1}}{1 - q} $

This is the standard sum for the given geometric series, assuming $q \neq 1$.

Analyzing the Options and Provided Answer

We can manipulate the standard result $S = \frac{1 - q^{r+1}}{1 - q}$ algebraically. For instance, multiplying the numerator and denominator by -1 gives an equivalent form:

$ S = \frac{-(1 - q^{r+1})}{-(1 - q)} = \frac{q^{r+1} - 1}{q - 1} $

Another way to express the sum is:

$ S = \frac{1}{1-q} - \frac{q^{r+1}}{1-q} $

The provided correct answer text is:

$ \frac{1}{q-1} - \frac{q^{r+1}}{1-q} $

This expression corresponds to Option 1. Let's analyze this expression further:

$ \frac{1}{q-1} - \frac{q^{r+1}}{1-q} $

To combine these terms using a common denominator $(q-1)$, we can rewrite the second term:

$ \frac{q^{r+1}}{1-q} = \frac{q^{r+1}}{-(q-1)} = -\frac{q^{r+1}}{q-1} $

Substituting this back:

$ \frac{1}{q-1} - \left(-\frac{q^{r+1}}{q-1}\right) = \frac{1}{q-1} + \frac{q^{r+1}}{q-1} = \frac{1 + q^{r+1}}{q-1} $

The standard sum derived is $S = \frac{1 - q^{r+1}}{1 - q}$. The expression provided as the correct answer is $\frac{1}{q-1} - \frac{q^{r+1}}{1-q}$, which matches Option 1.