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1 1 . 2 + 1 2 . 3 + 1 3 . 4 + . . . + 1 N ( N + 1 ) = N N + 1

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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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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 4 | Page 27

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