Advertisement Remove all ads

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

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

Advertisement Remove all ads

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 .

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

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

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×