Prove the statement by the Principle of Mathematical Induction: $1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$ for all natural numbers $n$.

(N/A) $P(n): 1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$
Step $1$: For $n=1$,
$L.H.S. = 1+2 = 3$
$R.H.S. = 2^{1+1}-1 = 2^{2}-1 = 4-1 = 3$
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,
$1+2+2^{2}+\ldots+2^{k}=2^{k+1}-1 \quad \dots(i)$
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1): 1+2+2^{2}+\ldots+2^{k}+2^{k+1} = 2^{(k+1)+1}-1$
$L.H.S. = (1+2+2^{2}+\ldots+2^{k}) + 2^{k+1}$
Using equation $(i)$:
$L.H.S. = (2^{k+1}-1) + 2^{k+1}$
$= 2 \times 2^{k+1} - 1$
$= 2^{k+2} - 1$
$= 2^{(k+1)+1} - 1 = R.H.S.$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.