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Question
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
Solution
Let P(n) be the given statement.
Now,
\[P(n): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
\[\text{ Step } 1: \]
\[ P(1) = \frac{1}{2} = 1 - \frac{1}{2^1}\]
\[\text{ Thus, P(1) is true .} \]
\[\text{ Step 2: } \]
\[\text{ Suppose P(m) is true .} \]
\[\text{ Then,} \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} = 1 - \frac{1}{2^m}\]
\[\text{ To show: P(m + 1) is true whenever P(m) is true } . \]
\[\text{ That is, } \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{{}^{m + 1}}} = 1 - \frac{1}{2^{m + 1}}\]
\[\text{ Now, P(m) is true } . \]
\[\text{ Thus, we have: } \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} = 1 - \frac{1}{2^m}\]
\[ \Rightarrow \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} + \frac{1}{2^{m + 1}} = 1 - \frac{1}{2^m} + \frac{1}{2^{m + 1}} \left[ \text{ Adding } \frac{1}{2^{m + 1}} \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = 1 - \frac{1}{2^m} + \frac{1}{2^m . 2} = 1 - \frac{1}{2^{{}^m}}\left( 1 - \frac{1}{2} \right) = 1 - \frac{1}{2^{m + 1}}\]
Thus , P ( m + 1) is true
$By the principle of mathematical induction, P(n) is true for all n ∈ N .
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