1+q+q**2+...+q**n+q**n+1=(1-q**n+2)/(1-q)

The equation you provided is almost correct, but there seems to be a small mistake in the exponent. Let’s analyze it step by step.

The left-hand side of the equation represents a **geometric series**:

\[
S = 1 + q + q^2 + \cdots + q^{n} + q^{n+1}
\]

This is the sum of a geometric series with the first term \(a = 1\), the common ratio \(q\), and a total of \(n + 2\) terms.

The sum of a finite geometric series is given by the formula:

\[
S = \frac{1 - q^{n+2}}{1 - q}, \quad \text{for} \quad q \neq 1.
\]

So the sum on the left-hand side should be:

\[
1 + q + q^2 + \cdots + q^{n+1} = \frac{1 - q^{n+2}}{1 - q}.
\]

But the right-hand side of your equation is:

\[
\frac{1 - q^{n+2}}{1 - q}.
\]

It looks like you had a slight typo where you wrote \(n+2\) as \(n+1\) on the left-hand side of the equation.

Thus, the correct equation is:

\[
1 + q + q^2 + \cdots + q^{n+1} = \frac{1 - q^{n+2}}{1 - q}.
\]

### Would you like more details on how this geometric series formula is derived or any other clarifications?

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Here are five related questions:
1. What happens to the series sum if \(q = 1\)?
2. How would the sum change if we extend the series to infinity?
3. Can you find the sum of a geometric series if \(q = -1\)?
4. How does the formula change if the common ratio is less than 1?
5. How does a geometric series relate to compound interest calculations?

**Tip:** Geometric series are widely used in fields like economics, physics, and computer science due to their predictable summation properties!

1 + q + q**2 + ... + q**n + q**n + 1 = (1 - q**n + 2) / (1 - q)

Explore the solution to the geometric series sum 1 + q + q^2 + ... + q^(n+1) and learn how to correctly apply the geometric series formula. Understand the process of identifying and correcting common mistakes in series summation, suitable for students in Grades 10-12.

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