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Question
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1 . 3 + 2 . 4 + 3 . 5 + . . . + n . (n + 2) = \frac{1}{6}n(n + 1)(2n + 7)\]
\[\text{ Step } 1: \]
\[P(1) = 1 . 3 = 3 = \frac{1}{6} \times 1(1 + 1)(2 \times 1 + 7)\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2:} \]
\[\text{ Let P(m) be true . } \]
\[\text{ Then, } \]
\[1 . 3 + 2 . 4 + . . . + m . (m + 2) = \frac{1}{6}m(m + 1)(2m + 7)\]
\[\text{ To prove: P(m + 1) is true . } \]
\[\text{ That is, } \]
\[1 . 3 + 2 . 4 + . . . + (m + 1)(m + 3) = \frac{1}{6}(m + 1)(m + 2)(2m + 9)\]
\[P(m) \text{ is equal to } 1 . 3 + 2 . 4 + . . . + m(m + 2) = \frac{1}{6}m(m + 1)(2m + 7) . \]
\[\text{ Thus, we have: } \]
\[1 . 3 + 2 . 4 + . . . + m(m + 2) + (m + 1)(m + 3) = \frac{1}{6}m(m + 1)(2m + 7) + (m + 1)(m + 3) \left[ \text{ Adding } (m + 1)(m + 3)\text{ to both sides } \right]\]
\[ \Rightarrow 1 . 3 + 2 . 4 + . . . + (m + 1)(m + 3) = \frac{1}{6}(m + 1)\left[ 2 m^2 + 7m + 6m + 18 \right]\]
\[ = \frac{1}{6}(m + 1)(2 m^2 + 13m + 18)\]
\[ = \frac{1}{6}(m + 1)(2m + 9)(m + 2)\]
\[\text{ Thus, P(m + 1) is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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