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Chapter 12: Q 42. (page 825)

In Problems 39– 44, find each sum.
$\sum_{k = 1}^{n} 4 \cdot 3^{k - 1}$

Short Answer

Expert verified

The sum of the given sequence is $2 \cdot (3^{n} - 1)$

Step by step solution

01

Step 1. Write the given information.

The given sequence is $\sum_{k = 1}^{n} 4 \cdot 3^{k - 1}$

02

Step 2. Find the sum.

$a_{1} = 4, r = 3$

The formula for the sum is $s_{n} = \frac{a_{1} (r^{n} - 1)}{r - 1}$

$s_{n} = \frac{4 (3^{n} - 1)}{3 - 1} = \frac{4 (3^{n} - 1)}{2} = 2 \cdot (3^{n} - 1)$

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Most popular questions from this chapter

In Problems 37–50, a sequence is defined recursively. Write down the first five terms.

$a_{1} = 3; a_{n} = 4 - a_{n - 1}$

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

$\sum_{k = 1}^{\infty} \frac{1}{2} 3^{k - 1}$

In Problems 29–36, the given pattern continues. Write down the nth term of a sequence $\{a_{n}\}$ suggested by the pattern.

$\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81}, . . .$

In Problems 37–50, a sequence is defined recursively. Write down the first five terms.

$a_{1} = 1; a_{n} = n - a_{n - 1}$

In Problems 11–16, evaluate each factorial expression.

$\frac{5! 8!}{3!}$

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