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Question
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1 + 3 + 5 + . . . + (2n - 1) = n^2 \]
\[\text{ Step } 1: \]
\[P(1) = 1 = 1^2 \]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step } 2 : \]
\[\text{ Let P(m) be true . } \]
\[\text{ Then }, \]
\[1 + 3 + 5 + . . . + (2m - 1) = m^2 \]
\[\text{ To prove:} P(m + 1) \text{ is true } . \]
\[i . e . , \]
\[1 + 3 + 5 + . . . + \left\{ 2\left( m + 1 \right) - 1 \right\} = \left( m + 1 \right)^2 \]
\[ \Rightarrow 1 + 3 + 5 + . . . + \left( 2m + 1 \right) = \left( m + 1 \right)^2 \]
\[\text{ Now, we have: } \]
\[1 + 3 + 5 + . . . + (2m - 1) = m^2 \]
\[ \Rightarrow 1 + 3 + . . . + (2m - 1) + (2m + 1) = m^2 + 2m + 1 \left[ \text{ Adding } 2m + 1 \text{ to both sides } \right]\]
\[ \Rightarrow 1 + 3 + 5 + . . . + (2m + 1) = (m + 1 )^2 \]
\[\text{ Hence, P(m + 1) is true} . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N . \]
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