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# 32n+7 is Divisible by 8 for All N ∈ N. - Mathematics

32n+7 is divisible by 8 for all n ∈ N.

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#### Solution

Let P(n) be the given statement.
Now,

$P(n): 5^{2n} - 1 \text{ is divisible by 24 for all n} \in N .$
$\text{ Step } 1:$
$P(1) = 5^2 - 1 = 25 - 1 = 24$
$\text{ It is divisible by } 24 .$
$\text{ Thus, P(1) is true } .$
$\text{ Step} 2:$
$\text{ Let P(m) be true .}$
$\text{ Then, 5^{2m} - 1 is divisible by 24 .}$
$\text{ Now, let} 5^{2m} - 1 = 24\lambda, \text{ where } \lambda \in N .$
$\text{ We need to show that P(m + 1) is true whenever P(m) is true } .$
$\text{ Now,}$
$P(m + 1) = 5^{2m + 2} - 1$
$= 5^{2m} 5^2 - 1$
$= 25(24\lambda + 1) - 1$
$= 600\lambda + 24$
$= 24(25\lambda + 1)$
$\text{ It is divisible by 24 } .$
$\text{ Thus, P(m + 1) is true } .$
$\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 20 | Page 28
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