\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Binomial Theorem

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The Binomial Theorem is a technique to expand binomial terms efficiently without having to multiply out and collect terms. There is some notation that you need to learn but it is not a hard technique and will save you time in your calculus work.

Binomial Theorem

\[ (a+b)^n = \sum_{k=0}^{n}{ \frac{n!}{(n-k)!k!}a^{n-k}b^k } \]

Understanding The Notation

1. A binomial term looks like \(a+b\), i.e. there are exactly 2 terms, a and b, added together and raised to some power.
2. The exponent n must be an integer greater than 1.
3. If you need a refresher on factorials, you can review them on the factorials page.
4. You may also see alternate notation for the factorial term like these.
\[ \frac{n!}{(n-k)!k!} = \left( \begin{array}{c} n \\ k \end{array} \right) = {}_{n}C_{k} \] The notation \( {}_{n}C_{k} \) is read 'n choose k'.
5. If you are not familiar with summation (sigma) notation, here is a good video that explains how it works.

Khan Academy - Sequences and Series (part 1) [9min-48secs]

In this video he introduces sequences, series and sigma notation. He does a pretty good job explaining the basics required for understanding series.

About 3 minutes into the video, he says something that is correct in this context but not in general. He says that you can add numbers in any order and get the same answer. This is true for a finite list of numbers, like he is doing here, but it is NOT always true if you have an infinite list of numbers. You will probably run across this case in your textbook while you are studying this topic.

video by Khan Academy

Understanding and Using The Theorem

Here is a great video explaining, step-by-step, how to use the binomial theorem with an example. When applying the binomial theorem, we sometimes say that we are finding the binomial expansion.

NancyPi - How to Use the Binomial Theorem [19min-58secs]

video by NancyPi

Pascal's Triangle

One way to calculate the factorial term is to use Pascal's Triangle. Here is a video explaining this with a couple of examples.

PatrickJMT - Pascal's Triangle and the Binomial Coefficients [5min-40secs]

video by PatrickJMT

Practice

Unless otherwise instructed, find the binomial expansion using the Binomial Theorem.

\((x+2)^5\)

Problem Statement

Find the binomial expansion of \((x+2)^5\) using the Binomial Theorem.

Solution

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\((x+y)^5\)

Problem Statement

Find the binomial expansion of \((x+y)^5\) using the Binomial Theorem.

Solution

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Find the binomial expansion of \((a+b)^5\) and use the result to expand \((x+1)^5\).

Problem Statement

Find the binomial expansion of \((a+b)^5\) and use the result to expand \((x+1)^5\).

Solution

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video by PatrickJMT

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\((a+5b)^3\)

Problem Statement

Find the binomial expansion of \((a+5b)^3\) using the Binomial Theorem.

Solution

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\((x-4)^5\)

Problem Statement

Find the binomial expansion of \((x-4)^5\) using the Binomial Theorem.

Solution

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\((2x-3)^4\)

Problem Statement

Find the binomial expansion of \((2x-3)^4\) using the Binomial Theorem.

Solution

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\((3x-y)^3\)

Problem Statement

Find the binomial expansion of \((3x-y)^3\) using the Binomial Theorem.

Solution

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\((2t-s)^5\)

Problem Statement

Find the binomial expansion of \((2t-s)^5\) using the Binomial Theorem.

Solution

2496 video

video by Brian McLogan

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\((3x^2-5y)^4\)

Problem Statement

Find the binomial expansion of \((3x^2-5y)^4\) using the Binomial Theorem.

Solution

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Find the third term of the binomial expansion of \((2x+3)^4\).

Problem Statement

Find the third term of the binomial expansion of \((2x+3)^4\).

Solution

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Determine the third term of the expansion of \((3x+1)^8\).

Problem Statement

Determine the third term of the expansion of \((3x+1)^8\).

Solution

2495 video

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Find the fifth term of the expansion of \((2x-3y)^7\).

Problem Statement

Find the fifth term of the expansion of \((2x-3y)^7\).

Solution

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Really UNDERSTAND Precalculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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