Binomial Theorem

To prove this theorem, a method called Mathematical Induction will be used. First the small cases need to be examined. Consider that

n = 1 .

(x + y)^{1} = x + y

Since

_{1} C_{0}

and

_{1} C_{1}

are both equal to

1,

the theorem holds for this case. Now consider

n = 2 .

(x + y)^{2} = x^{2} + 2 x y + y^{2}

On this case,

_{2} C_{0}

and

_{2} C_{2}

are equal

1

and

_{2} C_{1}

is equal to

2,

making this case also true. Following this, it can be assumed that the theorem holds for a natural number

k .

To prove the theorem, it needs to be shown that the theorem holds for the value of

k + 1 .

This can be done by multiplying the case above by

(x + y) .

(x + y)^{k + 1} = (x + y) (x + y)^{k}

To expand the binomial, multiply

(x + y)

by the binomial expansion of

(x + y)^{k} .

This has to be done carefully, as distributing

x

and

y

modify the terms on the binomial expansion.

Examining the resulting terms closely, it can be noted that there are terms that can be factored together. For example, consider the following.

_{k} C_{1} x^{k} y^{1} +_{k} C_{0} x^{k} y^{1}_{k} C_{2} x^{k - 1} y^{2} +_{k - 1} C_{1} x^{k - 1} y^{2} = (_{k} C_{0} +_{k} C_{1}) x^{k} y^{1} = (_{k} C_{1} +_{k} C_{2}) x^{k - 1} y^{2}

Using these examples, it is possible to find a way to write these terms. Let

r

be a natural number from

1

k .

Using this variable, factored terms can be written as follows.

(_{k} C_{r - 1} +_{k} C_{r}) x^{k - r + 1} y^{r}

Now the sum between parenthesis will be examined to see if it can be simplified. The expressions for the combinations will be used to do so.

_{k} C_{r - 1} +_{k} C_{r}

▼

Simplify

SubstituteExpressions

Substitute expressions

\frac{k !}{( k - ( r - 1 ) ) ! ( r - 1 ) !} + \frac{k !}{( k - ( r ) ) ! ( r ) !}

FactorOut

Factor out $k!$

k! (\frac{1}{( k - ( r - 1 ) ) ! ( r - 1 ) !} + \frac{1}{( k - ( r ) ) ! ( r ) !})

DistrNegSignSwap

$- (b - a) = a - b$

k! (\frac{1}{( k + 1 - r ) ! ( r - 1 ) !} + \frac{1}{( k - r ) ! ( r ) !})

ExpandFrac

$\frac{a}{b} = \frac{a \cdot r}{b \cdot r}$

k! (\frac{r}{( k + 1 - r ) ! ( r - 1 ) ! r} + \frac{1}{( k - r ) ! ( r ) !})

$(r - 1)! r = r!$

k! (\frac{r}{( k + 1 - r ) ! ( r ) !} + \frac{1}{( k - r ) ! ( r ) !})

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( k + 1 - r )}{b \cdot ( k + 1 - r )}$

k! (\frac{r}{( k + 1 - r ) ! ( r ) !} + \frac{k + 1 - r}{( k - r ) ! ( r ) ! ( k + 1 - r )})

$(k - r)! (k + 1 - r) = (k + 1 - r)!$

k! (\frac{r}{( k + 1 - r ) ! ( r ) !} + \frac{k + 1 - r}{( k + 1 - r ) ! ( r ) !})

AddFrac

Add fractions

k! (\frac{k + 1}{( k + 1 - r ) ! ( r ) !})

Multiply

\frac{k ! ( k + 1 )}{( k + 1 - r ) ! ( r ) !}

$k! (k + 1) = (k + 1)!$

\frac{( k + 1 ) !}{( k + 1 - r ) ! ( r ) !}

SubstituteExpressions

Substitute expressions

_{k + 1} C_{r}

Considering this result, the repeating terms can be rewritten to simplify the binomial expansion.

Looking at the binomial expansion, it can be noted that the repeated terms are written like the theorem. The remaining terms $x^{k + 1}$ and $y^{k + 1}$ can be rewritten considering the values of $_{k} C_{0} x^{k + 1}$ and · $_{k} C_{k} x^{0} y^{k + 1} .$

Combination	Formula	Simplify
$_{k} C_{0}$	$\frac{k !}{( k - 0 ) ! 0 !}$	$1$
$_{k} C_{k}$	$\frac{k !}{( k - k ) ! k !}$	$1$
$_{k + 1} C_{0}$	$\frac{( k + 1 ) !}{( k + 1 - 0 ) ! 0 !}$	$1$
$_{k + 1} C_{k + 1}$	$\frac{( k + 1 ) !}{( k + 1 - ( k + 1 ) ) ! ( k + 1 ) !}$	$1$

Since of the values on the table are equal to $1,$ these can be interchanged to rewrite the expression one last time.

Therefore, the result holds for in the case that $n = k + 1,$ which finish the proof.

{{ article.displayTitle }}

Binomial Theorem

Proof

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}