1 3 . 7 + 1 7 . 11 + 1 11 . 5 + . . . + 1 ( 4 N − 1 ) ( 4 N + 3 ) = N 3 ( 4 N + 3 )

प्रश्न

\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]

उत्तर

Let P(n) be the given statement.
Now,

\[P(n) = \frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 15} + . . . + \frac{{}^1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]

\[\text{ Step} 1: \]

\[P(1) = \frac{1}{3 . 7} = \frac{1}{21} = \frac{1}{3(4 + 3)}\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step } 2: \]

\[\text{ Let P(m) is true} . \]

\[\text{ Then} , \]

\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + . . . + \frac{1}{(4m - 1)(4m + 3)} = \frac{m}{3(4m + 3)}\]

\[\text{ To prove: P(m + 1) is true .} \]

\[\text{ That is} , \]

\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + . . . + \frac{1}{(4m + 3)(4m + 7)} = \frac{m + 1}{3(4m + 7)}\]

\[Now, \]

\[P(m) = \frac{1}{3 . 7} + \frac{1}{7 . 11} + . . . + \frac{1}{(4m - 1)(4m + 3)} = \frac{m}{3(4m + 3)}\]

\[ \Rightarrow \frac{1}{3 . 7} + \frac{1}{7 . 11} + . . . + \frac{1}{(4m - 1)(4m + 3)} + \frac{1}{(4m + 3)(4m + 7)} = \frac{m}{3(4m + 3)} + \frac{1}{(4m + 3)(4m + 7)} \left[ \text{ Adding } \frac{1}{(4m + 3)(4m + 7)} \text{ to both sides } \right]\]

\[ \Rightarrow \frac{1}{3 . 7} + \frac{1}{7 . 11} + . . . + \frac{1}{(4m + 3)(4m + 7)} = \frac{4 m^2 + 7m + 3}{3(4m + 3)(4m + 7)} = \frac{(4m + 3)(m + 1)}{3(4m + 3)(4m + 7)} = \frac{m + 1}{3(4m + 7)}\]

\[\text{ Thus, P(m + 1) is true } . \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]

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Principle of Mathematical Induction

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 9 Q 8 Q 10

अध्याय 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २७]

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आर.डी. शर्मा Mathematics [English] Class 11

अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 9 | पृष्ठ २७

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Principle of Mathematical Induction

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