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Question
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
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Solution
Let P(n) be the given statement.
Now,
\[P(n) = \frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{1 . 2} = \frac{1}{2} = \frac{1}{1 + 1}\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step } 2: \]
\[\text{ Let P(m) be true } \]
\[\text{ Then,} \]
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{m(m + 1)} = \frac{m}{m + 1}\]
\[\text{ We shall now prove that P(m + 1) is true } . \]
\[i . e . , \]
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{(m + 1)(m + 2)} = \frac{m + 1}{m + 2}\]
\[ \text{ Now } , \]
\[P(m) = \frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{m(m + 1)} = \frac{m}{m + 1}\]
\[ \Rightarrow \frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{m(m + 1)} + \frac{1}{(m + 1)(m + 2)} = \frac{m}{m + 1} + \frac{1}{(m + 1)(m + 2)} \left[ \text{ Adding } \frac{1}{(m + 1)(m + 2)} \text{ to both sides} \right]\]
\[ \Rightarrow \frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{(m + 1)(m + 2)} = \frac{m^2 + 2m + 1}{(m + 1)(m + 2)} = \frac{(m + 1 )^2}{(m + 1)(m + 2)} = \frac{m + 1}{m + 2}\]
\[\text{ Therefore, P(m + 1) is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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