12-14-18-12 n -1-12 n

Let P(n) be the given statement. Now, P(n): #160;12+14+18+...+12n=1 12nStep #160;1: #160;P(1)=12=1 121Thus, #160;P(1) #160;is #160;true.Step #160; ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

12+14+18+…+12...

Question

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

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Solution

Let P(n) be the given statement.
Now,
$P (n) : \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}} Step 1 : P (1) = \frac{1}{2} = 1 - \frac{1}{2^{1}} Thus, P (1) is true . Step 2 : Suppose P (m) is true . Then, \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{m}} = 1 - \frac{1}{2^{m}} T o s h o w : P (m + 1) is true whenever P (m) is true . That is, \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{^{m + 1}}} = 1 - \frac{1}{2^{m + 1}} N o w, P (m) is true . T h u s, w e h a v e : \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{m}} = 1 - \frac{1}{2^{m}} \Rightarrow \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{m}} + \frac{1}{2^{m + 1}} = 1 - \frac{1}{2^{m}} + \frac{1}{2^{m + 1}} [Adding \frac{1}{2^{m + 1}} to both sides] \Rightarrow P (m + 1) = 1 - \frac{1}{2^{m}} + \frac{1}{2^{m} . 2} = 1 - \frac{1}{2^{^{m}}} (1 - \frac{1}{2}) = 1 - \frac{1}{2^{m + 1}} T h u s, P (m + 1) is true .$
$B y the p rinciple of m athematical i nduction, P (n) is true for a l l n \in N .$

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12+14+18+...+12n=1-12n

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