मराठी

Prove that 1 N + 1 + 1 N + 2 + . . . + 1 2 N > 13 24 , for All Natural Numbers N > 1 - Mathematics

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प्रश्न

\[\text{ Prove that }  \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers } n > 1 .\]

 

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उत्तर

\[\text{ Let p} \left( n \right): \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers } n > 1 . \]
\[\text{ Step I: For n }  = 2, \]
\[LHS = \frac{1}{2 + 1} + \frac{1}{2 \times 2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} = \frac{14}{24} > \frac{13}{24} = RHS\]
\[As, LHS > RHS\]
\[\text{ So, it is true for n } = 2 . \]
\[\text{ Step II: For n }  = k, \]
\[\text{ Let } p\left( k \right): \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} > \frac{13}{24}, \text{ be true for some natural numbers }  k > 1 . \]
\[\text{ Step III: For n } = k + 1, \]
\[p\left( k + 1 \right) = \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} + \frac{1}{2k + 1}\]
\[ > \frac{13}{24} + \frac{1}{2k + 1} \left( \text{ Using step } II \right)\]
\[ > \frac{13}{24}\]
\[i . e . p\left( k + 1 \right) > \frac{13}{24}\]
\[\text{ So, it is also true for n }  = k + 1 . \]
\[\text{ Hence,}  p\left( n \right): \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers}  n > 1 .\]

 

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पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 12 Mathematical Induction
Exercise 12.2 | Q 41 | पृष्ठ २९

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