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(1) Use mathematical Induction to prove the formd ar sum_(i=0)^n q^i=(q^n+1-1)/(q-1) with q in mathbb(R)

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(1) Use mathematical Induction to prove the formd ar sum_(i=0)^n q^i=(q^n+1-1)/(q-1) with q in mathbb(R)

(1) Use mathematical Induction to prove the formd ar sum_(i=0)^n q^i=(q^n+1-1)/(q-1) with q in mathbb(R)

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VivianMestre · Tutor por 5 anos

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To prove the formula \( \sum_{i=0}^{n} q^{i}=\frac{q^{n+1}-1}{q-1} \) using mathematical induction, we will follow these steps:<br /><br />1. **Base Case**: Show that the formula holds for \( n = 0 \).<br />2. **Inductive Step**: Assume that the formula holds for some arbitrary positive integer \( k \), and then show that it holds for \( k + 1 \).<br /><br />### Step 1: Base Case<br /><br />For \( n = 0 \):<br />\[ \sum_{i=0}^{0} q^i = q^0 = 1 \]<br /><br />The right-hand side of the formula is:<br />\[ \frac{q^{0+1} - 1}{q - 1} = \frac{q - 1}{q - 1} = 1 \]<br /><br />Thus, the base case holds true.<br /><br />### Step 2: Inductive Step<br /><br />Assume that the formula holds for some arbitrary positive integer \( k \):<br />\[ \sum_{i=0}^{k} q^i = \frac{q^{k+1} - 1}{q - 1} \]<br /><br />We need to show that the formula holds for \( k + 1 \):<br />\[ \sum_{i=0}^{k+1} q^i = \frac{q^{(k+1)+1} - 1}{q - 1} \]<br /><br />Consider the left-hand side:<br />\[ \sum_{i=0}^{k+1} q^i = \sum_{i=0}^{k} q^i + q^{k+1} \]<br /><br />By the inductive hypothesis:<br />\[ \sum_{i=0}^{k} q^i = \frac{q^{k+1} - 1}{q - 1} \]<br /><br />So,<br />\[ \sum_{i=0}^{k+1} q^i = \frac{q^{k+1} - 1}{q - 1} + q^{k+1} \]<br /><br />Combine the terms over a common denominator:<br />\[ \sum_{i=0}^{k+1} q^i = \frac{q^{k+1} - 1 + q^{k+1}(q - 1)}{q - 1} \]<br />\[ = \frac{q^{k+1} - 1 + q^{k+2} - q^{k+1}}{q - 1} \]<br />\[ = \frac{q^{k+2} - 1}{q - 1} \]<br /><br />Thus, the formula holds for \( k + 1 \).<br /><br />### Conclusion<br /><br />By mathematical induction, the formula \( \sum_{i=0}^{n} q^i = \frac{q^{n+1} - 1}{q - 1} \) is proven to be true for all \( n \in \mathbb{N} \).
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