1 + 3 + 5 + … + (2n − 1) =
Question:

$1+3+5+\ldots+(2 n-1)=n^{2}$ i.e., the sum of first $n$ odd natural numbers is $n^{2}$.

Solution:

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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