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A + Ar + Ar2 + ... + Arn−1 = a ( R N − 1 R − 1 ) , R ≠ 1

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Question

a + ar + ar2 + ... + arn−1 =  \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]

 
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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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Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

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R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 17 | Page 27

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