Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on a few proofs and am missing how this algebra works....

So, how does one expand $(k+1)^3\,$? Can I use FOIL? What does it expand to?

And how to expand $(k+1)^5\,$?

Thanks!

share|improve this question
1  
look up Pascal's Triangle, and Binomial Theorem. Also no, you can't use FOIL, at very least you must use distributive property. –  Joseph Skelton Jan 22 '13 at 20:59

4 Answers 4

up vote 12 down vote accepted

Check out the entry Binomial Theorem in Wikipedia.

Putting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\; (k+1)^5, \;$ and $\,(k + 1)^n\,$ for any non-negative integer $n$.

FOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$

One can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$

But for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.

I'll include an animation and image of "Pascal's Triangle" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:

$\quad\quad\quad\quad$enter image description here $\quad\quad\quad\quad$enter image description here

$$\text{Each number in the triangle is the sum of the two directly above it.}$$


To see how this "plays out" in the expansion of $(x + 1)^n,\;0 \le n \le 6$:

$$(x + 1)^0 = \color{blue}{\bf{1}}$$ $$(x + 1)^1 = \color{blue}{\bf{1}}\cdot x +\color{blue}{\bf{1}}$$ $$(x + 1)^2 = \color{blue}{\bf{1}}\cdot x^2 + \color{blue}{\bf{2}}x + \color{blue}{\bf{1}}$$ $$(x+1)^3 = \color{blue}{\bf{1}}\cdot x^3 + \color{blue}{\bf{3}}x^2 + \color{blue}{\bf{3}}x + \color{blue}{\bf{1}}$$ $$(x+1)^4 = \color{blue}{\bf{1}}\cdot x^4 + \color{blue}{\bf{4}} x^3+ \color{blue}{\bf{6}}x^2 + \color{blue}{\bf{4}}x +\color{blue}{\bf{1}}$$ $$(x+1)^5 = \color{blue}{\bf{1}}\cdot x^5 + \color{blue}{\bf{5}}x^4 + \color{blue}{\bf{10}} x^3 + \color{blue}{\bf{10}} x^2 + \color{blue}{\bf{5}}x + \color{blue}{\bf{1}}$$ $$(x + 1)^6 = \color{blue}{\bf{1}}\cdot x^6 + \color{blue}{\bf{6}}x^5 +\color{blue}{\bf{15}}x^4 + \color{blue}{\bf{20}}x^3 +\color{blue}{\bf{15}}x^2 + \color{blue}{\bf{6}}x + \color{blue}{\bf{1}}$$ $${\bf{\vdots}}$$

share|improve this answer
    
Image source: Wikipedia, found at link to "Pascal's Triangle". –  amWhy Jan 22 '13 at 22:17
1  
Nice animated GIF! The answer is good, too :-) (+1) –  robjohn Jan 22 '13 at 22:53
    
Thanks, @robjohn! It says more concisely, in animation, what would take a bit of explaining to write! –  amWhy Jan 22 '13 at 22:56
    
@amWhy: It is unbelievable!! Great!! I cannot say anything. Nothing such this can describe the problem soooo well. +10! –  B. S. Jan 23 '13 at 15:50
    
This is incredible! Perfect explanation. Thank you! –  user56763 Jan 23 '13 at 16:44

You can use "FOIL" twice. You should get $$ k^3+3k^2+3k+1. $$ More generally, $$ (a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3. $$

Please don't vacilate between lower-case $k$ and capital $K$ in mathematical notation. Pick one and stick to it. Mathematical notation is case sensitive. Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.

But it would take a while to use FOIL to get $$ (a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9. $$ That's one reason to be aware of the binomial theorem, which explains the pattern.

share|improve this answer

Before you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that $$ (k+1)^3=(k+1)(k+1)(k+1) $$ We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have $$ (k+1)(k+1)=k^2+2k+1 $$ We're two-thirds of the way to the answer. Now we have $$ (k+1)^3=(k+1)(k^2+2k+1) $$ and by the distributive property, namely that $(a+b)c=ac+bc$, we have $$ \begin{align} (k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\ &=(k^3+2k^2+k)+(k^2+2k+1)\\ &=k^3+3k^2+3k+1 \end{align} $$ This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.

share|improve this answer
    
Thanks, makes sense. –  user56763 Jan 23 '13 at 16:45
    
@ user56763: From what Rick Decker posted you can see that $(k+1)^3 = (k+1)(k+1)(k+1)$ can be expanded in the following way. Write down all possible 3-fold products of the form $ABC,$ where $A$ is a choice of one of the terms in the first $(k+1)$ factor, $B$ is a choice of one of the terms in the second $(k+1)$ factor, and $C$ is a choice of one of the terms in the third $(k+1)$ factor. Then add the 8 products. Thus, among the things you'll add will be $(k)(k)(1)$ and $(1)(k)(k)$ and $(k)(1)(k).$ Note that there are 3 of these with exactly two $k$'s and exactly one $1,$ so you get $3k^2$ ... –  Dave L. Renfro Jan 28 '13 at 19:30

What you're looking for is the binomial theorem, where y = 1.

share|improve this answer
    
This is very scant. Something more would be good: some explanation or at least a link to an article. Of course, an answer should be more than simply a link. Look at the other answers for examples. –  robjohn Jan 22 '13 at 22:56
    
Yeah, I was going to, but was having trouble typing out the math for a more exhaustive answer. –  MITjanitor Jan 23 '13 at 2:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.