# 1.6 The geometric progression and the binomial theorem

 Page 1 / 1
The binomial theorem is introduced, the existences of nth roots of real numbers is explored, the binomial coefficient is defined, and a theorem providing a formula for the sum of a geometric progression is included.

There are two special algebraic identities that hold in $R$ (in fact in any field $F$ whatsoever) that we emphasize. They are bothproved by mathematical induction. The first is the formula for the sum of a geometric progression.

## Geometric progression

Let $x$ be a real number, and let $n$ be a natural number. Then,

1. If $x\ne 1,$ then
$\sum _{j=0}^{n}{x}^{j}=\frac{1-{x}^{n+1}}{1-x}.$
2. If $x=1,$ then
$\sum _{j=0}^{n}{x}^{j}=n+1.$

The second claim is clear, since there are $n+1$ summands and each is equal to 1.

We prove the first claim by induction. Thus, if $n=1,$ then the assertion is true, since

$\sum _{j=0}^{1}{x}^{j}={x}^{0}+{x}^{1}=1+x=\left(1+x\right)\frac{1-x}{1-x}=\frac{1-{x}^{2}}{1-x}.$

Now, supposing that the assertion is true for the natural number $k,$ i.e., that

$\sum _{j=0}^{k}{x}^{j}=\frac{1-{x}^{k+1}}{1-x},$

let us show that the assertion holds for the natural number $k+1.$ Thus

$\begin{array}{ccc}\hfill \sum _{j=0}^{k+1}{x}^{j}& =& \sum _{j=0}^{k}{x}^{j}+{x}^{k+1}\hfill \\ & =& \frac{1-{x}^{k+1}}{1-x}+{x}^{k+1}\hfill \\ & =& \frac{1-{x}^{k+1}+{x}^{k+1}-{x}^{k+2}}{1-x}\hfill \\ & =& \frac{1-{x}^{k+1+1}}{1-x},\hfill \end{array}$

which completes the proof.

The second algebraic formula we wish to emphasize is the Binomial Theorem. Before stating it, we must introduce some useful notation.

Let $n$ be a natural number. As earlier in this chapter, we define $n!$ as follows:

$n!=n×\left(n-1\right)×\left(n-2\right)×...×2×1.$

For later notational convenience, we also define $0!$ to be 1.

If $k$ is any integer for which $0\le k\le n,$ we define the binomial coefficient $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ by

$\left(\genfrac{}{}{0pt}{}{n}{k}\right)=\frac{n!}{k!\left(n-k\right)!}=\frac{n×\left(n-1\right)×\left(n-2\right)×...×\left(n-k+1\right)}{k!}.$
1. Prove that $\left(\genfrac{}{}{0pt}{}{n}{0}\right)=1,\phantom{\rule{0.166667em}{0ex}}\left(\genfrac{}{}{0pt}{}{n}{1}\right)=n\phantom{\rule{0.166667em}{0ex}}$ and $\left(\genfrac{}{}{0pt}{}{n}{n}\right)=1.$
2. Prove that
$\left(\genfrac{}{}{0pt}{}{n}{k}\right)\le \frac{2{n}^{k}}{{2}^{k}}$
for all natural numbers $n$ and all integers $0\le k\le n.$
3. Prove that
$\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)=\left(\genfrac{}{}{0pt}{}{n}{k}\right)+\left(\genfrac{}{}{0pt}{}{n}{k-1}\right)$
for all natural numbers $n$ and all integers $1\le k\le n.$

If $x,y\in R$ and $n$ is a natural number, then

${\left(x+y\right)}^{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){x}^{k}{y}^{n-k}.$

We shall prove this theorem by induction. If $n=1,$ then the assertion is true, for ${\left(x+y\right)}^{1}=x+y$ and

$\sum _{k=0}^{1}\left(\genfrac{}{}{0pt}{}{1}{k}\right){x}^{k}{y}^{1-k}=\left(\genfrac{}{}{0pt}{}{1}{0}\right){x}^{0}{y}^{1}+\left(\genfrac{}{}{0pt}{}{1}{1}\right){x}^{1}{y}^{0}=x+y.$

Now, assume that the assertion holds for the natural number $j;$ i.e.,

${\left(x+y\right)}^{j}=\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j-k},$

and let us prove that the assertion holds for the natural number $j+1.$ We will make use of part (c) of [link] . We have that

$\begin{array}{ccc}\hfill {\left(x+y\right)}^{j+1}& =& \left(x+y\right){\left(x+y\right)}^{j}\hfill \\ & =& \left(x+y\right)\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j-k}\hfill \\ & =& x\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j-k}+y\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j-k}\hfill \\ & =& \sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k+1}{y}^{j-k}+\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1-k}\hfill \\ & =& \sum _{k=0}^{j-1}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k+1}{y}^{j-k}+\left(\genfrac{}{}{0pt}{}{j}{j}\right){x}^{j+1}{y}^{0}\hfill \\ & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}& +\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1-k}+\left(\genfrac{}{}{0pt}{}{j}{0}\right){x}^{0}{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k-1}\right){x}^{k}{y}^{j+1-k}\hfill \\ & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}& +\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1-k}+{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}\left(\left(\genfrac{}{}{0pt}{}{j}{k-1}\right)+\left(\genfrac{}{}{0pt}{}{j}{k}\right)\right){x}^{k}{y}^{j+1-k}+{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1-k}+{y}^{j+1}\hfill \\ & =& \left(\genfrac{}{}{0pt}{}{j+1}{j+1}\right){x}^{j+1}{y}^{0}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1-k}+\left(\genfrac{}{}{0pt}{}{j+1}{0}\right){x}^{0}{y}^{j+1}\hfill \\ & =& \sum _{k=0}^{j+1}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1-k},\hfill \end{array}$

which shows that the assertion of the theorem holds for the natural number $j+1.$ This completes the proof.

The next exercise is valid in any ordered field, but, since we are mainly interested in the order field $R,$ we state everything in terms of that field.

1. If $x$ and $y$ are positive real numbers, and if $n$ and $k$ are natural numbers with $k\le n,$ show that ${\left(x+y\right)}^{n}\ge \left(\genfrac{}{}{0pt}{}{n}{k}\right){x}^{k}{y}^{n-k}.$
2. For any positive real number $x$ and natural number $n,$ show that ${\left(1+x\right)}^{n}\ge 1+nx.$
3. For any real number $x>-1$ and natural number $n,$ prove that ${\left(1+x\right)}^{n}\ge 1+nx.$ HINT: Do not try to use the binomial theorem as in part (b); it won't work because the terms are not all positive; prove this directly by induction.

There is one more important algebraic identity, which again can be proved by induction.It is actually just a corollary of the geometric progression formula.

If $x,y\in R$ and $n$ is a natural number, then

${x}^{n}-{y}^{n}=\left(x-y\right)\left(\sum _{j=0}^{n-1}{x}^{j}{y}^{n-1-j}.$

If $n=1$ the theorem is clear. Suppose it holds for a natural number $k,$ and let us prove the identity for the natural number $k+1.$ We have

$\begin{array}{ccc}\hfill {x}^{k+1}-{y}^{k+1}& =& {x}^{k+1}-{x}^{k}y+{x}^{k}y-{y}^{k+1}\hfill \\ & =& \left(x-y\right){x}^{k}+y\left({x}^{k}-{y}^{k}\right)\hfill \\ & =& \left(x-y\right){x}^{k}+y\left(x-y\right)\left(\sum _{j=0}^{k-1}{x}^{j}{y}^{k-1-j}\right)\hfill \\ & =& \left(x-y\right){x}^{k}+\left(x-y\right)\left(\sum _{j=0}^{k-1}{x}^{j}{y}^{k-j}\hfill \\ & =& \left(x-y\right)\left({x}^{k}{y}^{k-k}+\sum _{j=0}^{k-1}{x}^{j}{y}^{k-j}\right)\hfill \\ & =& \left(x-y\right)\left(\sum _{j=0}^{k}{x}^{j}{y}^{k-j}\right)\hfill \end{array}$

, which shows that the assertion holds for the natural number $k+1.$ So, by induction, the theorem is proved.

Let $x$ and $y$ be real numbers.

1. Let $n$ be an odd natural number; i.e., $n=2k+1$ for some natural number $k.$ Show that
${x}^{n}+{y}^{n}=\left(x+y\right)\left(\sum _{j=0}^{n-1}{\left(-1\right)}^{j}{x}^{j}{y}^{n-1-j}.$
HINT: Write ${x}^{n}+{y}^{n}={x}^{n}-{\left(-y\right)}^{n}.$
2. Show that ${x}^{2}+{y}^{2}$ can not be factored into a product of the form $\left(ax+by\right)\left(cx+dy\right)$ for any choices of real numbers $a,b,c,$ and $d.$

Using the Binomial Theorem together with the preceding theorem, we may now investigate the existence of $n$ th roots of real numbers. This next theorem is definitely not valid in any ordered field, forit again depends on the completeness property.

Let $n$ be a natural number and let $x$ be a positive real number. Then there exists a unique positive real number $y$ such that ${y}^{n}=x;$ i.e., $x$ has a unique positive $n$ th root.

Note first that if $0\le t then ${t}^{n}<{s}^{n}.$ (To see this, argue by induction, and use part (e) of [link] .) Using this, we mimic the proof of [link] . Thus, let $S$ be the set of all positive real numbers $t$ for which ${t}^{n}\le x.$ Then $S$ is nonempty and bounded above. Indeed, if $x\ge 1,$ then $1\in S,$ while if $x<1,$ then $x$ itself is in $S.$ Therefore, $S$ is nonempty. Also, using part (b) of [link] , we see that $1+\left(x/n\right)$ is an upper bound for $S.$ For, if $t>1+x/n,$ then

${t}^{n}>{\left(1+\left(x/n\right)\right)}^{n}\ge 1+n\left(x/n\right)>x.$

Now let $y=supS,$ and let us show that ${y}^{n}=x.$ We rule out the other two possibilities. First, if ${y}^{n}>x,$ let $ϵ$ be the positive number ${y}^{n}-x,$ and define ${ϵ}^{\text{'}}$ to be the positive number $ϵ/\left(n{y}^{n-1}\right).$ Then, using [link] , choose $t\in S$ so that $y-{ϵ}^{\text{'}} ( [link] is where the completeness of the ordered field $R$ is crucial.) We have

$\begin{array}{ccc}\hfill ϵ& =& {y}^{n}-x\hfill \\ & =& {y}^{n}-{t}^{n}+{t}^{n}-x\hfill \\ & \le & {y}^{n}-{t}^{n}\hfill \\ & =& \left(y-t\right)\left(\sum _{j=0}^{n-1}{y}^{j}{t}^{n-1-j}\right)\hfill \\ & \le & \left(y-t\right)\left(\sum _{j=0}^{n-1}{y}^{j}{y}^{n-1-j}\right)\hfill \\ & =& \left(y-t\right)\left(\sum _{j=0}^{n-1}{y}^{n-1}\hfill \\ & <& {ϵ}^{\text{'}}n{y}^{n-1}\hfill \\ & =& ϵ,\hfill \end{array}$

and this is a contradiction. Therefore, ${y}^{n}$ is not greater than $x.$

Now, if ${y}^{n} let $ϵ$ be the positive number $x-{y}^{n},$ and choose a $\delta >0$ such that $\delta <1$ and $\delta <ϵ/{\left(y+1\right)}^{n}.$ Then, using the Binomial Theorem, we have that

$\begin{array}{ccc}\hfill {\left(y+\delta \right)}^{n}& =& \sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta }^{n-k}\hfill \\ & =& {y}^{n}+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta }^{n-k}\hfill \\ & =& {y}^{n}+\delta \sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta }^{n-1-k}\hfill \\ & <& {y}^{n}+\delta \sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{1}^{n-k}\hfill \\ & =& {y}^{n}+\delta {\left(y+1\right)}^{n}\hfill \\ & =& x-ϵ+\delta {\left(y+1\right)}^{n}\hfill \\ & <& x-ϵ+ϵ\hfill \\ & =& x,\hfill \end{array}$

implying that $y+\delta \in S.$ But this is a contradiction, since $y=supS.$ Therefore, ${y}^{n}$ is not less than $x,$ and so ${y}^{n}=x.$

We have shown the existence of a positive $n$ th root of $x.$ To see the uniqueness, suppose $y$ and ${y}^{\text{'}}$ are two positive $n$ th roots of $x.$ Then

$\begin{array}{ccc}\hfill 0& =& {y}^{n}-{{y}^{\text{'}}}^{n}\hfill \\ & =& \left(y-{y}^{\text{'}}\right)\left(\sum _{j=0}^{n-1}{y}^{j}{{y}^{\text{'}}}^{n-j-1},\hfill \end{array}$

which implies that either $y-{y}^{\text{'}}=0$ or ${\sum }_{j=0}^{n-1}{y}^{j}{{y}^{\text{'}}}^{n-j-1}=0.$ Since this latter sum consists of positive terms, it cannot be 0, whence $y={y}^{\text{'}}.$ This shows that there is but one positive $n$ th root of $x,$ and the theorem is proved.

1. Show that if $n=2k$ is an even natural number, then every positive real number has exactly two distinct $n$ th roots.
2. If $n=2k+1$ is an odd natural number, show that every real number has exactly one $n$ th root.
3. If $n$ is a natural number greater than 1, prove that there is no rational number whose $n$ th power equals 2, i.e., the $n$ th root of 2 is not a rational number.

anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!