Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) :
The sum of n terms of the Progression $$1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+.... is \dfrac{2^{(n-1)}-1}{2^{(n-1)}}$$

Reason (R) :
The sum of a geometric series having n terms is given by $$S_{n} = \dfrac{a(1-r^{n})}{1-r}$$, where a is the $$1^{st}$$ term and r is the common ratio.

Reason (R) states that the sum of a geometric series having n terms is given by $$S_{n} = \dfrac{a(1-r^{n})}{1-r}$$, where a is the $$1^{st}$$ term and r is the common ratio. We know that this is the sum of geometric series, thus the reason is true.

Now Assertion (A) states that the sum of n terms of the Progression $$1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+.... is \dfrac{2^{(n-1)}-1}{2^{(n-1)}}$$

Here a = 1 and r = 1/2. We will substitute these values in the formula given in (R).

$$S_n=\dfrac{1(1-\left(\frac{1}{2}\right)^n)}{1-\frac{1}{2}}$$

$$S_n=\dfrac{2\left(2^n-1\right)}{2^n}$$

$$S_n=\dfrac{2^n-1}{2^{n-1}}$$

This is not equal to $$\dfrac{2^{(n-1)}-1}{2^{(n-1)}}$$. Thus, the assertion is false.