Question

evaluate:
find the first term.
\\\sum_{n=1}^{5} 6(2)^{n-1}\\
\\s_n = \frac{?(1 - \quad^{\quad})}{1 - \quad} \quad a = ?\\
remember: for a geometric series, \\s_n = \frac{a(1-r^n)}{1-r}\\

Explanation:

Identify the general term of the series

The given series is written in sigma notation:

$$ \sum_{n=1}^{5} 6(2)^{n-1} $$

The general term \( a_n \) of this series is:

$$ a_n = 6(2)^{n-1} $$

Find the first term

To find the first term \( a \), substitute \( n = 1 \) into the general term:

$$ a = a_1 = 6(2)^{1-1} $$

$$ a = 6(2)^0 $$

$$ a = 6(1) = 6 $$

Identify the common ratio and number of terms

The general term of a geometric sequence is \( a_n = a(r)^{n-1} \).
Comparing this to \( 6(2)^{n-1} \):

• First term \( a = 6 \)
• Common ratio \( r = 2 \)
• Number of terms \( n = 5 \)

Substitute values into the sum formula

Using the formula for the sum of a geometric series:

$$ S_n = \frac{a(1-r^n)}{1-r} $$

Substitute \( a = 6 \), \( r = 2 \), and \( n = 5 \):

$$ S_5 = \frac{6(1-2^5)}{1-2} $$

Comparing this to the template \( S_n = \frac{[?](1-\Box^{\Box})}{1-\Box} \), we fill in the blanks.

Answer:

Evaluate:
Find the first term.

$$ \sum_{n=1}^{5} 6(2)^{n-1} $$

\( S_n = \) <blank>\(\frac{6(1-2^5)}{1-2}\)</blank> \( a = \) <blank>\(6\)</blank>