Prove the statement by the Principle of Mathematical Induction :
$1+5+9+\ldots+(4 n-3)=n(2 n-1)$ for all natural numbers $n$.

(N/A) $P(n): 1+5+9+\ldots+(4 n-3)=n(2 n-1)$
For $n=1, \quad L.H.S.=1$
$R$.$H$.$S$. $=1(2(1)-1) = 1(1) = 1$
$\therefore L.H.S. = R.H.S.$
$\therefore P(1)$ is true.
Assume that $P(k)$ is true for some $k \in \mathbb{N}$.
$P(k): 1+5+9+\ldots+(4 k-3)=k(2 k-1) \quad \ldots(i)$
For $n=k+1$,we need to show $P(k+1)$ is true:
$L.H.S. = [1+5+9+\ldots+(4 k-3)] + (4(k+1)-3)$
$= k(2 k-1) + (4 k+4-3) \quad (\text{using } (i))$
$= 2 k^{2}-k+4 k+1$
$= 2 k^{2}+3 k+1$
$= 2 k^{2}+2 k+k+1$
$= 2k(k+1) + 1(k+1)$
$= (k+1)(2 k+1)$
$= (k+1)[2(k+1)-1] = R.H.S.$
$\therefore P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.