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Question
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
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Solution
Let P(n) be the given statement.
Now,
\[P(n) = 2 + 5 + 8 + . . . + (3n - 1) = \frac{1}{2}n(3n + 1)\]
\[\text{ Step} 1: \]
\[P(1) = 2 = \frac{1}{2} \times 1(3 + 1) \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step} 2: \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then, } \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ To prove: P(m + 1) is true } . \]
\[\text{ That is, } \]
\[2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(m + 1)(3m + 4)\]
\[\text{ P(m) is equal to:} \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ Thus, we have:} \]
\[2 + 5 + 8 + . . . + (3m - 1) + (3m + 2) = \frac{1}{2}m(3m + 1) + (3m + 2) \left[ \text{ Adding } (3m + 2) \text{ to both sides } ] \right]\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3 m^2 + m + 6m + 4) = \frac{1}{2}(3 m^2 + 7m + 4)\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3m + 4)(m + 1)\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
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