# 1 2 + 1 4 + 1 8 + . . . + 1 2 N = 1 − 1 2 N - Mathematics

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}$

#### 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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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 15 | Page 27