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Question
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
Solution
Let P(n) be the given statement.
Now,
\[P(n) = a + ar + a r^2 + . . . + a r^{n - 1} = a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
\[\text{ Step } 1: \]
\[P(1) = a = a\left( \frac{r^1 - 1}{r - 1} \right)\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2} : \]
\[\text{ Suppose P(m) is true } . \]
\[\text{ Then } , \]
\[a + ar + a r^2 + . . . + a r^{m - 1} = a\left( \frac{r^m - 1}{r - 1} \right), r \neq 1\]
\[\text{ To show: P(m + 1) is true whenever P(m) is true } . \]
\[\text{ That is, } \]
\[a + ar + a r^2 + . . . + a r^m = a\left( \frac{r^{m + 1} - 1}{r - 1} \right), r \neq 1\]
\[\text{ We know that P(m) is true} . \]
\[\text{ Thus, we have: } \]
\[a + ar + a r^2 + . . . + a r^{m - 1} = a\left( \frac{r^m - 1}{r - 1} \right)\]
\[ \Rightarrow a + ar + a r^2 + . . . + a r^{m - 1} + a r^m = a\left( \frac{r^m - 1}{r - 1} \right) + a r^m \left[ \text{ Adding } a r^m \text{ to both sides} \right]\]
\[ \Rightarrow P(m + 1) = a\left( \frac{r^m - 1 + r . r^m - r^m}{r - 1} \right)\]
\[ \Rightarrow P(m + 1) = a\left( \frac{r^{m + 1} - 1}{r - 1} \right), r \neq 1\]
\[\text{ Thus, P(m + 1) is true . } \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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