/*! 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 30 Use the Binomial Theorem to expa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 30

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (a+2 b)^{6} $$

Short Answer

Expert verified
(a+2b)^6 equals to a^6 + 12a^5b + 60a^4b^2 + 160a^3b^3 + 240a^2b^4 + 192a*b^5 + 64b^6

Step by step solution

01

Identify p, q and n

p corresponds to the term 'a', q corresponds to the term '2b', and n corresponds to '6'.
02

Use the Binomial Theorem

(a+2b)^6 = 危 (from k=0 to 6) [6Ck * a^(6-k) * (2b)^k]. This is based on the Binomial Theorem.
03

Expand the expression

Start off with k=0 and increment k by 1 each step until reaching 6. Calculate each term independently, then add these terms together to get the simplified form of the expansion.
04

Solve for each term

Term for k=0: 6C0 * a^(6-0) * (2b)^0= 1*a^6*1=a^6.\nTerm for k=1: 6C1 * a^(6-1) * (2b)^1 = 6*a^5*(2b) = 12a^5b.\nTerm for k=2: 6C2*a^(6-2)*(2b)^2 = 15*a^4*(2*2)b^2 = 60a^4b^2.\nTerm for k=3: 6C3*a^(6-3)*(2b)^3 = 20*a^3*(2*2*2)b^3 = 160a^3b^3. \nTerm for k=4: 6C4*a^(6-4)*(2b)^4 = 15*a^2*(2*2*2*2)b^4 = 240a^2b^4. \nTerm for k=5: 6C5*a^(6-5)*(2b)^5 = 6*a*(2*2*2*2*2)b^5 = 192a*b^5. \nTerm for k=6: 6C6*a^(6-6)*(2b)^6 = 1*(2*2*2*2*2*2)b^6 = 64b^6.
05

Add all terms

Add all these terms together to get the final expanded form: (a+2b)^6 = a^6 + 12a^5b + 60a^4b^2 + 160a^3b^3 + 240a^2b^4 + 192a*b^5 + 64b^6

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 value of an annuity to solve Exercises 77鈥�84. Round answers to the nearest dollar. To offer scholarship funds to children of employees, a company invests \(\$ 10,000\) at the end of every three months in an annuity that pays \(10.5 \%\) compounded quarterly. a. How much will the company have in scholarship funds at the end of ten years? b. Find the interest.

What is the difference between a geometric sequence and an infinite geometric series?

The table shows the population of Texas for 2000 and 2010 with estimates given by the U.S. Census Bureau for 2001 through 2009 \(\begin{array}{llllll}{\text { Year }} & {2000} & {2001} & {2002} & {2003} & {2004} \\ \hline \text { Population } & {20.85} & {21.27} & {21.70} & {22.13} & {22.57} & {23.02} \\\ \hline\end{array}\) \(\begin{array}{llllll}{\text { Year }} & {2006} & {2007} & {2008} & {2009} & {2010} \\ \hline \text { Population } & {23.48} & {23.95} & {24.43} & {24.92} & {25.15} \\ {\text { in millions }} & {23.48} & {23.95} & {24.43} & {24.92} & {25.15}\end{array}\) a. Divide the population for each year by the population in the preceding year. Round to two decimal places and show that Texas has a population increase that is approximately geometric. b. Write the general term of the geometric sequence modeling Texas's population, in millions, \(n\) years after 1999 c. Use your model from part (b) to project Texas's population, in millions, for the year \(2020 .\) Round to two decimal places.

Mega Millions is a multi-state lottery played in most U.S. states. As of this writing, the top cash prize was \(\$ 656\) million, going to three lucky winners in three states. Players pick five different numbers from 1 to 56 and one number from 1 to \(46 .\) Use this information to solve Exercises \(27-30 .\) Express all probabilities as fractions. A player wins a minimum award of \(\$ 150\) by correctly matching three numbers drawn from white balls \((1 \text { through } 56\) ) and matching the number on the gold Mega Ball \((1 \text { through } 46)\) What is the probability of winning this consolation prize?

Explaining the Concepts Write a probability word problem whose answer is one of the following fractions: \(\frac{1}{6}\) or \(\frac{1}{4}\) or \(\frac{1}{3}\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Decision Maths

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.