Let the given statement be P(n), i.e.,
P (n) :

1

2

+

1

4

+

1

8

+ ... +

1

2n

= 1 -

1

2n

First we prove that the statement is true for n = 1
P (1) :

1

2

= 1 -

1

2

, which is true
Assume P(k) is true for some positive integer k, i.e.,

1

2

+

1

4

+

1

8

+ ... +

1

2k

= 1 -

1

2k

... (i)
We shall now prove that P(k + 1) is also true.
For this we have to prove that

1

2

+

1

4

+

1

8

+ ... +

1

2k

+

1

2k+1

= 1 -

1

2k+1

L.H.S. =

1

2

+

1

4

+

1

8

+ ... +

1

2k

+

1

2k+1

= 1 -

1

2k

−

1

2k+1

[From (i)]
= 1 - [

1

2k

−

1

2k+1

] = 1 -

1

2k

[1−

1

2

] = 1 -

1

2k

.

1

2

= 1 -

1

2k+1

= R.H.S.
Thus P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.