Chapter 7: Q. 34 (page 592)

In Exercises 31–36 provide the first five terms of the given sequence. Unless specified, assume that the first term has index 1.
$\{\frac{(- 1)^{k - 1} x^{2 k}}{2 k!}\}$

Short Answer

Expert verified

The first five terms are $\frac{x^{2}}{2!}, - \frac{x^{4}}{4!}, \frac{x^{6}}{6!}, - \frac{x^{8}}{8!}, \frac{x^{10}}{10!}$ .

Step by step solution

Step 1. Given information.

The given sequence is $\{\frac{(- 1)^{k - 1} x^{2 k}}{2 k!}\}$ .

Step 2. First term.

The general sequence is $a_{k} = \frac{(- 1)^{k - 1} x^{2 k}}{2 k!}$ .

When $k = 1,$

$a_{1} = \frac{(- 1)^{1 - 1} x^{2 (1)}}{2 (1)!} = \frac{x^{2}}{2!}$

Step 3. Remaining terms.

Now, put

$k = 2, a_{2} = \frac{(- 1)^{2 - 1} x^{2 (2)}}{2 (2)!} a_{2} = \frac{- x^{4}}{4!} k = 3, a_{3} = \frac{(- 1)^{3 - 1} x^{2 (3)}}{2 (3)!} = \frac{x^{6}}{6!} k = 4, a_{4} = \frac{(- 1)^{4 - 1} x^{2 (4)}}{2 (4)!} = \frac{- x^{8}}{8!} k = 5, a_{5} = \frac{(- 1)^{5 - 1} x^{2 (5)}}{2 (5)!} = \frac{x^{10}}{10!}$

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In Exercises 31–36 provide the first five terms of the given sequence. Unless specified, assume that the first term has index 1.
$\{\frac{(- 1)^{k - 1} x^{2 k}}{2 k!}\}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. First term.

Step 3. Remaining terms.

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In Exercises 31–36 provide the first five terms of the given sequence. Unless specified, assume that the first term has index 1.(-1)k-1x2k2k!

Short Answer

Step by step solution

Step 1. Given information.

Step 2. First term.

Step 3. Remaining terms.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Mechanics Maths

Statistics

Probability and Statistics

Logic and Functions

Pure Maths

Study anywhere. Anytime. Across all devices.

In Exercises 31–36 provide the first five terms of the given sequence. Unless specified, assume that the first term has index 1.
$\{\frac{(- 1)^{k - 1} x^{2 k}}{2 k!}\}$