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