$1+3+5+\ldots+(2 n-1)=n^{2}$ i.e., the sum of first $n$ odd natural numbers is $n^{2}$.
Let P(n) be the given statement.
Now,
$P(n)=1+3+5+\ldots+(2 n-1)=n^{2}$
Step 1:
$P(1)=1=1^{2}$
Hence, $P(1)$ is true.
Step 2;
Let $P(m)$ be true.
Then:
$1+3+5+\ldots+(2 m-1)=m^{2}$
To prove : $P(m+1)$ is true.
i. e.,
$1+3+5+\ldots+\{2(m+1)-1\}=(m+1)^{2}$
$\Rightarrow 1+3+5+\ldots+(2 m+1)=(m+1)^{2}$
Now, we have;
$1+3+5+\ldots+(2 m-1)=m^{2}$
$\Rightarrow 1+3+\ldots+(2 m-1)+(2 m+1)=m^{2}+2 m+1 \quad$ [Adding $2 m+1$ to both sides]
$\Rightarrow 1+3+5+\ldots+(2 m+1)=(m+1)^{2}$
Hence, $P(m+1)$ is true.
By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in \mathrm{N}$.
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