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1 2 . 5 + 1 5 . 8 + 1 8 . 11 + . . . + 1 ( 3 N − 1 ) ( 3 N + 2 ) = N 6 N + 4

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Question

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

 

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Solution

Let P(n) be the given statement.
Now, 

\[P(n) = \frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

\[\text{ Step }  1: \]

\[P(1) = \frac{1}{2 . 5} = \frac{1}{10} = \frac{1}{6 + 4}\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2:}  \]

\[\text{ Let P(m) be true .}  \]

\[\text{ Then,}  \]

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3m - 1)(3m + 2)} = \frac{m}{6m + 4}\]

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

\[i . e . , \]

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m + 2)(3m + 5)} = \frac{m + 1}{6m + 10}\]

\[\text{ Thus, we have }: \]

\[ \frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3m - 1)(3m + 2)} = \frac{m}{6m + 4}\]

\[ \Rightarrow \frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m - 1)(3m + 2)} + \frac{1}{(3m + 2)(3m + 5)} = \frac{m}{6m + 4} + \frac{1}{(3m + 2)(3m + 5)} \left[ \text{ Adding } \frac{1}{(3m + 2)(3m + 5)} \text{ to both sides } \right]\]

\[ \Rightarrow \frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m + 2)(3m + 5)} = \frac{3 m^2 + 5m + 2}{2(3m + 2)(3m + 5)} = \frac{(3m + 2)(m + 1)}{2(3m + 2)(3m + 5)} = \frac{m + 1}{6m + 10}\]

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

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

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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 6 | Page 27

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