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Question
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
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Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1 + 3 + 3^2 + . . . + 3^{n - 1} = \frac{3^n - 1}{2}\]
\[\text{ Step 1} : \]
\[P(1) = 1 = \frac{3^1 - 1}{2} = \frac{2}{2} = 1\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2:} \]
\[\text{ Let P(m) is true .} \]
\[\text{ Then } , \]
\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]
\[\text{ We shall prove that P(m + 1) is true } . \]
\[\text{ That is, } \]
\[1 + 3 + 3^2 + . . . + 3^m = \frac{3^{m + 1} - 1}{2}\]
\[\text{ Now, we have: } \]
\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]
\[ \Rightarrow 1 + 3 + 3^2 + . . . + 3^{m - 1} + 3^m = \frac{3^m - 1}{2} + 3^m \left[ \text{ Adding } 3^m \text{ to both sides } \right]\]
\[ \Rightarrow 1 + 3 + 3^2 + . . . + 3^m = \frac{3^m - 1 + 2 \times 3^m}{2} = \frac{3^m (1 + 2) - 1}{2} = \frac{3^{m + 1} - 1}{2}\]
\[\text{ Hence, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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