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1 1 . 4 + 1 4 . 7 + 1 7 . 10 + . . . + 1 ( 3 N − 2 ) ( 3 N + 1 ) = N 3 N + 1 - Mathematics

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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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अध्याय 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २७]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 7 | पृष्ठ २७

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