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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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