Proof ∀n∈ℕ, n>1, ∀r∈ℝ\{1}, Σ(rᵏ)[k=0→(n-1)] = (rⁿ - 1)/(r-1)

Show that $\forall n \in \mathbb{N}_{>1}, \forall r \in \mathbb{R} \setminus \{1\}, 1 + r + r^2 + \cdots + r^{n-1} = \frac {r^n - 1}{r-1}$.

We are asked to prove the formula for the finite geometric series:

$$ \forall n \in \mathbb{N}_{>1}, \forall r \in \mathbb{R} \setminus \{1\}, \quad 1 + r + r^2 + \cdots + r^{n-1} = \frac{r^n - 1}{r - 1} $$

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Let’s define the left-hand side:

Let:

$$ S = 1 + r + r^2 + \cdots + r^{n-1} $$

This is a geometric series with:

• First term $a = 1$
• Common ratio $r$
• Number of terms $n$

We want to derive a closed-form expression for $S$.

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Step 1: Multiply both sides by $r$

$$ rS = r + r^2 + r^3 + \cdots + r^n $$

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Step 2: Subtract the two equations

Subtract the original series $S$ from $rS$:

$$ rS - S = (r + r^2 + \cdots + r^n) - (1 + r + r^2 + \cdots + r^{n-1}) $$

Almost all terms cancel out:

$$ rS - S = r^n - 1 $$

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Step 3: Factor out $S$

$$ S(r - 1) = r^n - 1 $$

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Step 4: Solve for $S$

Since $r \ne 1$, we can divide both sides:

$$ S = \frac{r^n - 1}{r - 1} $$

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Conclusion

Thus, we have shown that:

$$ \boxed{1 + r + r^2 + \cdots + r^{n-1} = \frac{r^n - 1}{r - 1}} \quad \text{for all } n \in \mathbb{N}_{>1}, \; r \in \mathbb{R} \setminus \{1\} $$

✔️ Proven.