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1.2 + 2.3 + 3.4 + ... + N (N + 1) = N ( N + 1 ) ( N + 2 ) 3

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Question

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 
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Solution

Let P(n) be the given statement.
Now,

\[P(n) = 1 . 2 + 2 . 3 + 3 . 4 + . . . + n(n + 1) = \frac{n(n + 1)(n + 2)}{3}\]

\[\text{ Step }  1: \]

\[P(1) = 1 . 2 = 2 = \frac{1(1 + 1)(1 + 2)}{3}\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step } 2: \]

\[\text{ Let P(m) be true } . \]

\[\text{ Then, } \]

\[1 . 2 + 2 . 3 + . . . + m(m + 1) = \frac{m(m + 1)(m + 2)}{3}\]

\[\text{ To prove: P(m + 1) is true .}  \]

\[\text{ That is,}  \]

\[1 . 2 + 2 . 3 + . . . + (m + 1)(m + 2) = \frac{(m + 1)(m + 2)(m + 3)}{3}\]

\[\text{ Now, P(m) is }  \]

\[1 . 2 + 2 . 3 + . . . + m(m + 1) = \frac{m(m + 1)(m + 2)}{3}\]

\[ \Rightarrow 1 . 2 + 2 . 3 + . . . + m(m + 1) + (m + 1)(m + 2) = \frac{m(m + 1)(m + 2)}{3} + (m + 1)(m + 2)\]

\[ \Rightarrow P(m + 1) = \frac{(m + 1)(m + 2)(m + 3)}{3}\]

\[\text{ Thus, P(m + 1) is true .}  \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n }  \in N . '\]

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Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

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R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 14 | Page 27

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