/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Write the first three terms in e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 37

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$\left(y^{3}-1\right)^{20}$$

Short Answer

Expert verified
The first three terms of the binomial expansion \[ (y^{3}-1)^{20} \] are \[ y^{60} - 20y^{57} + 190y^{54} \].

Step by step solution

01

First Term Calculation

The first term will consist of the first part of the binomial, \(y^{3}\), raised to the power of 20, and the second part of the binomial, -1, to the power of 0. Use Binomial Theorem to calculate: \[ \binom{20}{0} * (y^{3})^{20} * (-1)^0 \]. Simplifying this will give: \[ y^{60} \]
02

Second Term Calculation

The second term is obtained by reducing the power of the first part of the binomial by 1, and increasing the power of the second part by 1. Use Binomial theorem to calculate: \[ \binom{20}{1} * (y^{3})^{19} * (-1)^1 \]. Simplifying this will give: \[ -20y^{57} \]
03

Third Term Calculation

The third term is obtained by further reducing the power of the first part of the binomial by 1, and increasing the power of the second part by another 1. Use Binomial theorem to calculate: \[ \binom{20}{2} * (y^{3})^{18} * (-1)^2 \]. Simplifying this will give: \[ 190y^{54} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the formula for the sum of the first n terms of a geometric sequence to solve. A pendulum swings through an arc of 20 inches. On each successive swing, the length of the arc is \(90 \%\) of the previous length. $$\begin{aligned}&20, \quad\quad 0.9(20), \quad0.9^{2}(20), \quad 0.9^{3}(20), \ldots\\\&\begin{array}{|c|c|c|c|}\hline \text { 1st } & \text { 2nd } & \text { 3rd } & \text { 4th } \\\\\text { swing } & \text { swing } & \text { swing } & \text { swing } \\\\\hline\end{array}\end{aligned}$$ After 10 swings, what is the total length of the distance the pendulum has swung? Round to the nearest hundredth of an inch.

Use the formula for the sum of the first n terms of a geometric sequence to solve. You save \(\$ 1\) the first day of a month, \(\$ 2\) the second day, \(\$ 4\) the third day, continuing to double your savings each day. What will your total savings be for the first 15 days?

$$\text { Solve: } \frac{6}{x}+\frac{6}{x+2}=\frac{5}{2}$$ (Section 7.6, Example 3)

Find a general term, \(a_{n},\) for each sequence. More than one answer may be possible. $$\frac{4}{1}, \frac{9}{2}, \frac{16}{3}, \frac{25}{4}, \dots$$

Use the formula for the general term (the nth term) of a geometric sequence to solve. Suppose you save \(\$ 1\) the first day of a month, \(\$ 2\) the second day, \(\$ 4\) the third day, and so on. That is, each day you save twice as much as you did the day before. What will you put aside for savings on the thirtieth day of the month?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.