Q. 34 In Exercises 31鈥�36 provide the... [FREE SOLUTION]

Chapter 7: Q. 34 (page 592)

In Exercises 31鈥�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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91影视

In Exercises 31鈥�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.

One App. One Place for Learning.

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91影视

In Exercises 31鈥�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

Probability and Statistics

Pure Maths

Theoretical and Mathematical Physics

Applied Mathematics

Discrete Mathematics

Geometry

Study anywhere. Anytime. Across all devices.

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