Question

If $ a_n = 2^{n-1} $, where $ n = 1, 2, 3, ... $, then find $ \sum_{n=1}^{20} a_n $.

• A. \( 2^{20} - 1 \)
• B. \( 2^{21} - 1 \)
• C. \( 2^{19} - 1 \)
• D. \( 2^{20} \)

Hint:

For a geometric series, remember the formula \( S_N = \frac{a(r^N - 1)}{r - 1} \). This is useful for quickly calculating sums of terms in geometric progressions.

Answer 1

The Correct Option is B

We are given the series: \[ a_n = 2^{n-1}, \quad n = 1, 2, 3, \dots \] The sum of the first 20 terms is: \[ S_{20} = \sum_{n=1}^{20} 2^{n-1} \] This is a geometric series with the first term \( a = 1 \) (since \( 2^0 = 1 \)) and the common ratio \( r = 2 \). 
The sum of the first \( N \) terms of a geometric series is given by: \[ S_N = \frac{a(r^N - 1)}{r - 1} \] Substituting the values \( a = 1 \), \( r = 2 \), and \( N = 20 \): \[ S_{20} = \frac{2^20 - 1}{2 - 1} = 2^{20} - 1 \] 
Thus, the sum of the first 20 terms is \( 2^{21} - 1 \), which corresponds to option (B).