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प्रश्न
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
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उत्तर
Let P(n) be the given statement.
Now,
\[P(n) = \frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{1 . 4} = \frac{1}{4} = \frac{1}{3 \times 1 + 1}\]
\[\text{ Hence, P(1) is true } . \]
\[Step 2: \]
\[\text{ Let P(m) be true} . \]
\[i . e . , \]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} = \frac{m}{3m + 1}\]
\[\text{ To prove: P(m + 1) is true } . \]
\[i . e . , \]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m + 1)(3m + 4)} = \frac{m + 1}{3m + 4}\]
\[\text{ Now, } \]
\[P(m) = \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} = \frac{m}{3m + 1}\]
\[ \Rightarrow \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} + \frac{1}{(3m + 1)(3m + 4)} = \frac{m}{3m + 1} + \frac{1}{(3m + 1)(3m + 4)} \left[ \text{ Adding } \frac{1}{(3m + 1)(3m + 4)} \text{ to both sides} \right]\]
\[ \Rightarrow \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m + 1)(3m + 4)} = \frac{3 m^2 + 4m + 1}{(3m + 1)(3m + 4)} = \frac{(3m + 1)(m + 1)}{(3m + 1)(3m + 4)} = \frac{m + 1}{3m + 4}\]
\[\text{ Thus, P(m + 1) is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
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