Pascal's Triangle and The Binomial Theorem
Good job, mathronaut! We're nearly to the pinnacle of Polynomial Mountain. At the top we'll plant our Shmoop flag high, snap a few photos to make the folks back home jealous, and find Pascal's Triangle and the Binomial Theorem.
These are both ways to quickly multiply out a binomial that's being raised by an exponent. Like, say:
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3
And so on. We could keep going all day, but eventually the exponents would get so big that it would take us all day to simplify them. We need a better way, and luckily a 17th-century French mathematician has already found one.
Blaise Pascal was an interesting dude. He studied physics, philosophy, religion, and mathematics—with maybe just a little help from alien polynomials from a certain planet. He found a numerical pattern, called Pascal's Triangle, for quickly expanding a binomial like the ones above.
Check it out. The first row is one 1. Then we have two 1s. After that, things get interesting. The outsides of the triangle are always 1, but the insides are different. To find the number on the next row, add the two numbers above it together. See, 2 = 1 + 1; for the row under that, we have 1 (outer edge), 3 (2 + 1 from above), 3 (ditto), and 1 (the other edge). Every row is built from the row above it.
Pascal's Triangle gives us the coefficients for an expanded binomial of the form (a + b)n, where n is the row of the triangle. The Binomial Theorem tells us we can use these coefficients to find the entire expanded binomial, with a couple extra tricks thrown in.
What about the variables and their exponents, though? Keep your pants on; the Binomial Theorem has us covered. At the left side of the triangle, we have anb0 = an. As we move to the right, we subtract 1 from a's exponent and add it to b's. It's like b keeps stealing exponents from a, one at a time, until he's got them all. That sneaky devil.
Pascal doesn't tell us about the sign of each term, though. Luckily, our polynomial friends have us covered. We're getting covered a lot here, aren't we?
Anyway, when a binomial has a "+" sign, such as (a + b)2, all of the terms of the expansion are positive.
When we expand a binomial with a "–" sign, such as (a – b)5, the first term of the expansion is positive and the successive terms will alternate signs.
With all this help from Pascal and his good buddy the Binomial Theorem, we're ready to tackle a few problems.
Sample Problem
Expand (x – y)4.
Take a look at Pascal's triangle. From the fourth row, we know our coefficients will be 1, 4, 6, 4, and 1. That negative sign means that the first term of our expansion will be positive, and the following terms will alternate signs. The exponents will start at x4y0 and move to x3y1, etc.
(x – y)4
= 1x4 – 4x3y + 6x2y2 – 4xy3 + 1y4
= x4 – 4x3y + 6x2y2 – 4xy3 + y4
Wasn't that much easier than trying to multiply the expression out? We probably wouldn't have even tried, no matter how much the polynomials complained to us.
Sample Problem
Expand (m + 2n)5.
Have Pascal, will travel (all the way to the answer). Our coefficients are 1, 5, 10, 10, 5, 1. The 2n doesn't change anything; treat it like a single term.
1m5 + 5m4(2n) + 10m3(2n)2 + 10m2(2n)3 + 5m(2n)4 + 1(2n)5
= m5 + 10m4n + 40m3n2 + 80m2n3 + 80mn4 + 32n5
With Pascal on our side, we feel invincible. However, that isn't quite the case. We're still weak to bullets, being dipped in lava while a villain monologues at us, or a well-timed chocolate. We also have trouble dealing with very large exponents; for every row we add to Pascal's triangle, it takes longer and longer to find the next. We also can't afford to skip any rows, or the usher will escort us out.
Often times, when we have a binomial with a huge exponent, we only need one or a few terms. For those problems, we'll use a special formula. It allows us to find the coefficient of just one specific term without finding any of the others.
It looks pretty weird, but stay with us.
Whew, that thing looks beastly. Just dreadful. It has fractions and exponents and factorials. We haven't even mentioned yet that the formula doesn't even tell you the sign of the term, either.
To find the rth term of a binomial expansion (a + b)n, plug the terms into the formula.
If the binomial has a "+" sign, then all terms found using this formula are positive.
If the binomial has a "–" sign, then the term is negative if r is even. If r is odd, then the term is positive.
Maybe it won't be so bad once we solve a problem or two with it. Don't judge a book-sized formula by its cover, eh?
Sample Problem
Find the 12th term of (x – y)14.
That binomial would take forever to multiply out, and we'd make hundreds of mistakes while doing it anyway. We're convinced already; give us the new formula, please.
Let's see: r = 12 and n = 14, while a = x and b = y. The 12th term is even, so our term will be negative.
Remember, a factorial is a shortcut for writing a bunch of multiplication, like 3! = 3 × 2 × 1. That means we can cancel out a lot of that 14! with the 11! below it, since 14! = 14 × 13 × 12 × 11!
Okay, we were skeptical at first, but this wasn’t actually too bad. As long as we know the formula, it works out.
Sample Problem
Find the 7th term of (x + y)17.
This is it—the last problem before the summit of Polynomial Mountain. We've come a long way, so let's push just a little farther.
We have an r of 7 and an n of 17. The term will be positive because we've got a plus sign in our binomial, so all the terms are gonna be positive.
Just think of how many terms we'd have to add together to get a number like 12,376 out of Pascal's Triangle.
With that out of the way, we've made it to the top. We've conquered Polynomial Mountain by learning the polynomials' ways, living among them, and befriending them in their alien home. Until next time, mathronauts!