Find the sum of 10 terms of the progression : 2+4+8+…

Explanation

To find the sum of the first 10 terms of the progression \(2 + 4 + 8 + \ldots\), we first identify the type of progression we are dealing with. This series is a geometric progression (GP) where:

1. The first term \(A = 2\).
2. The common ratio \(R\) can be found by dividing the second term by the first term:
\[ R = \frac{4}{2} = 2. \]

Next, we need to find the sum of the first \(n\) terms of a geometric progression. The formula for the sum \(S_n\) of the first \(n\) terms of a geometric progression when the common ratio \(R\) is greater than 1 is given by:
\[ S_n = A \cdot \frac{R^n - 1}{R - 1}. \]

In this case, we want to find \(S_{10}\):
- Here, \(n = 10\), \(A = 2\), and \(R = 2\).

Substituting these values into the formula, we get:
\[ S_{10} = 2 \cdot \frac{2^{10} - 1}{2 - 1}. \]

Calculating \(2 - 1\):
\[ 2 - 1 = 1. \]
So the formula simplifies to:
\[ S_{10} = 2 \cdot (2^{10} - 1). \]

Now, we calculate \(2^{10}\):
\[ 2^{10} = 1024. \]
Thus,
\[ S_{10} = 2 \cdot (1024 - 1) = 2 \cdot 1023 = 2046. \]

Therefore, the sum of the first 10 terms of the progression is:
\[ \boxed{2046}. \]