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1 + 2 + 3 + ... + n = n ( n + 1 ) 2 i.e. the sum of the first n natural numbers is n ( n + 1 ) 2 .

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Question

1 + 2 + 3 + ... + n =  \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .

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Solution

Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n  =\[\frac{n(n + 1)}{2}\]

\[\text{ Step}  1: \]

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

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

\[\text{ Step } 2: \]

\[\text{ Let P(m) be true }  . \]

\[\text{ Then, } \]

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

\[\text{ We shall now prove that P(m + 1) is true }  . \]

\[i . e . , \]

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

\[\text{ Now,}  \]

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

\[ \Rightarrow 1 + 2 + . . . m + m + 1 = \frac{m(m + 1)}{2} + m + 1 \left[ \text{ Adding } \left( m + 1 \right) \text{ to both sides }  \right]\]

\[ = \frac{m^2 + m + 2m + 2}{2}\]

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

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

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 1 | Page 27

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