# 1.6 The geometric progression and the binomial theorem

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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.

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