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1 + 3 + 5 + ... + (2n − 1) = N2 I.E., the Sum of First N Odd Natural Numbers is N2.

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Question

1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

 
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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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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 5 | Page 27

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