Solution:

Let P(n) be the given statement.

Now,

$P(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

Step 1;

$P(1)=\frac{1}{2}=1-\frac{1}{2^{1}}$

Thus, $P(1)$ is true.

Step 2 :

 

Suppose $P(m)$ is true.

Then,

$\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}=1-\frac{1}{2^{m}}$

To show: $P(m+1)$ is true whenever $P(m)$ is true.

That is,

$\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m+1}}=1-\frac{1}{2^{m+1}}$

Now, $P(m)$ is true.

Thus, we have:

$\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}=1-\frac{1}{2^{m}}$

$\Rightarrow \frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}+\frac{1}{2^{m+1}}=1-\frac{1}{2^{m}}+\frac{1}{2^{m+1}} \quad$ Adding $\frac{1}{2^{m+1}}$ to both sides

$\Rightarrow P(m+1)=1-\frac{1}{2^{m}}+\frac{1}{2^{m} \cdot 2}=1-\frac{1}{2^{m}}\left(1-\frac{1}{2}\right)=1-\frac{1}{2^{m+1}}$

Thus, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.