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