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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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