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