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1 3 . 5 + 1 5 . 7 + 1 7 . 9 + . . . + 1 ( 2 N + 1 ) ( 2 N + 3 ) = N 3 ( 2 N + 3 )

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Question

\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]

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Solution

Let P(n) be the given statement.
Now,

\[P(n) = \frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]

\[\text{ Step 1} : \]

\[P(1) = \frac{1}{3 . 5} = \frac{1}{15} = \frac{1}{3(2 + 3)}\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2 } : \]

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

\[\text{ Then, } \]

\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2m + 1)(2m + 3)} = \frac{m}{3(2m + 3)}\] 

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

\[\text{ That is } , \]

\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2m + 3)(2m + 5)} = \frac{m + 1}{3(2m + 5)}\]

\[Now, \]

\[P(m) = \frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2m + 1)(2m + 3)} = \frac{m}{3(2m + 3)}\]

\[ \Rightarrow \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{(2m + 1)(2m + 3)} + \frac{1}{(2m + 3)(2m + 5)} = \frac{m}{3(2m + 3)} + \frac{1}{(2m + 3)(2m + 5)} \left[ \text{ Adding } \frac{1}{(2m + 3)(2m + 5)} \text{ to both sides }  \right]\]

\[ \Rightarrow \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{(2m + 3)(2m + 5)} = \frac{2 m^2 + 5m + 3}{3(2m + 3)(2m + 5)} = \frac{(2m + 3)(m + 1)}{3(2m + 3)(2m + 5)} = \frac{m + 1}{3\left( 2m + 5 \right)}\]

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

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

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Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

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R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 8 | Page 27

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