Question

If the sum of $1 + 2 + {2^2} + ....... + {2^{n - 1}}$ is 255, find the value of $n$?

Answer 1

In the given question we need to find the value of $n$. Here, we will use the concept of geometric progression to solve the question. We will first find the ratio, then substitute the value of first term and common ratio in the formula for G.P. to find the required value.

Formula used:
To find the sum of first n terms, denoted by ${S_n}$ of a geometric sequence we will use the formula, ${S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$ , where $n$ is the number of terms, $a$ is the first term and $r$ is the common ratio.

Complete step by step solution:
The given series is a geometric progression.
Now we will find the common ratio of the given series. Therefore, we get
Common ratio, $r$=$\dfrac{2}{1}$=$\dfrac{{{2^2}}}{2}$
Now we can say the given series is a G.P. with the first term $a = 1$ and common ratio $r = 2$.
We are also given that the sum of $n$ terms of a given series is 255.
We know that the sum of $n$ terms is given by ${S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$.
Now substituting ${S_n} = 255$ in the formula, we get
$\Rightarrow \dfrac{{a(1 - {r^n})}}{{1 - r}} = 255$
Substituting the value of $r = 2$ and $a = 1$, we get
$\Rightarrow \dfrac{{1(1 - {2^n})}}{{1 - 2}} = 255$
$\Rightarrow \dfrac{{1 - {2^n}}}{{ - 1}} = 255$
Multiplying both sides of the above equation by $- 1$,we get
$\Rightarrow \dfrac{{1 - {2^n}}}{{ - 1}} \times \left( { - 1} \right) = 255\left( { - 1} \right)$
$\Rightarrow 1 - {2^n} = - 255$= -255
Subtracting 1 from both sides, we get
$\Rightarrow 1 - {2^n} - 1 = - 255 - 1$
$\Rightarrow {2^n} = 256$
$\Rightarrow {2^n} = {2^8}$

Therefore, by comparing both sides of the above equation, we get
$\Rightarrow n = 8$

Note:
Geometric progression is a type of progression or a series in which the consecutive terms differ by a common ratio. It is different from the arithmetic progression. An arithmetic progression is also a type of progression or series in which the consecutive terms have a common difference.