Consider the given statement
$P(n): 1+2^1+2^2+\ldots+2^n=2^{n+1}-1$, for all natural numbers $n$.
For $\mathrm{n}=1$,
$$
\begin{aligned}
P(1): 1+2 &=2^{1+1}-1 \\
3 &=2^2-1 \\
3 &=4-1 \\
3 &=3, \text { which is true. }
\end{aligned}
$$
Now, let $P(n)$ is true for $n=k$.
So, $P(k): 1+2^1+2^2+\ldots+2^k=2^{k+1}-1$ is true.
Now, to prove $P(k+1)$ is true.
$$
\begin{aligned}
&P(k+1): 1+2^1+2^2+\ldots+2^k+2^{k+1} \\
&=2^{k+1}-1+2^{k+1} \\
&=2 \cdot 2^{k+1}-1 \\
&=2^{k+2}-1=2^{(k+1)+1}-1
\end{aligned}
$$
So, $P(k+1)$ is true, whenever $P(k)$ is true.
Hence $P(n)$ is true.