Evaluate the Summation sum from x=1 to 5 of 4^(x-1)
The problem is asking for the calculation of a summation, specifically the sum of powers of four, beginning with 4^(x-1) where x starts from 1 and increases by increments of 1 until it reaches 5. You are required to find the total value after adding all of these terms together.
$\sum_{x = 1}^{5} 4^{x - 1}$
Answer
Solution:
Step 1:
Write out the terms of the series for each integer value of \( x \) from 1 to 5.
\( 4^{1 - 1} + 4^{2 - 1} + 4^{3 - 1} + 4^{4 - 1} + 4^{5 - 1} \)
Step 2:
Calculate the sum of the series.
\( 1 + 4 + 16 + 64 + 256 = 341 \)
Knowledge Notes:
The problem involves evaluating a finite summation of terms in the form of \( 4^{x-1} \), where \( x \) ranges from 1 to 5. This is a geometric series because each term is a constant multiple of the previous term. The general form of a term in a geometric series is \( ar^{n-1} \), where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
In this series:
- \( a = 4^{1-1} = 4^0 = 1 \) (the first term)
- \( r = 4 \) (the common ratio, since each term is multiplied by 4 to get the next term)
The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \] However, since the series in question is short, we can simply evaluate each term and add them together directly.
The steps to solve the problem are:
Expand the series by substituting the values of \( x \) from 1 to 5 into the term \( 4^{x-1} \).
Simplify the expanded series by calculating the value of each term and then summing them up to get the final result.
The result of the summation is \( 341 \), which is the sum of the series when \( x \) ranges from 1 to 5.
