Binomial TheoremBinomial Theorem for Positive Integral Indices
Start with the expansions you already know:
Each line is the line above it multiplied by . Throughout this chapter the first term a is blue, the second term b is green, and the numbers in front of the terms — the coefficients — are purple.
The multiplying machine. Every press multiplies the expansion by and shows the working: everything times a, everything times b, and then the like terms added up. Multiply your way up to .
Look at the last press. The × a row is the old row of coefficients. The × b row is the same numbers, moved one place to the right. So each number in the new row is the sum of the two numbers stacked above it.
Three more things hold for every expansion you built.
(i) It has
(ii) From each term to the next, the power of
(iii) In every term, the powers of and add up to the
Now write down only the coefficients, one row for each index (Fig 7.1):
| Index | Coefficients |
|---|---|
| 0 | 1 |
| 1 | 1 1 |
| 2 | 1 2 1 |
| 3 | 1 3 3 1 |
| 4 | 1 4 6 4 1 |
Every row starts and ends with 1, and each number inside it is the sum of the two numbers above it — 1 + 1 makes the 2 in row 2, while 1 + 2 and 2 + 1 make the two 3s in row 3. With that one rule we can write as many rows as we like.
Pascal's Triangle
Let's keep going. Stack the rows so that every number sits between the two numbers that make it, and they grow into a triangle — Fig 7.2. Fill in rows 5 and 6. Tap a cell and the two numbers that feed it light up.
This array is called Pascal's triangle, after the French mathematician
Meru Prastara
The triangle was discovered again and again, long before anyone called it Pascal's. Pingala's Chhandas Shastra, a book about the rhythms of Sanskrit verse, laid out these numbers for powers up to 7 in about 200 BCE. The Chinese mathematician Zhu Shijie (Chu Shih-chieh) drew the triangle in 1303. In Germany, Michael Stifel gave the numbers their name, binomial coefficients, around 1544. Rafael Bombelli listed the coefficients of up to in 1572, and William Oughtred up to in 1631. Pascal's Traité du triangle arithmétique, printed in 1665 after his death, stated the theorem for whole-number powers in the form we use today.
Put these milestones in order, earliest at the top.
Nearly two thousand years separate the first milestone from the last, and every one of these mathematicians was looking at the same rows you have just filled in.
Expanding with a Row
A row of the triangle is all you need to expand a binomial. Take . Here plays the part of a, plays the part of b, and row 5 is 1, 5, 10, 10, 5, 1. Using (i), (ii) and (iii),
Now simplify one term at a time. Raise the number and the letter to the power:
and . So
Choosing from Brackets
Now try the same way. You would first have to write out every row of the triangle down to row 12 — and for bigger powers, even more. We want a way to jump straight to any row. The trick is to ask where the numbers come from.
means . To multiply it out, you pick a or b from each bracket, multiply your four picks together, and do this for every possible set of picks. Tap a bracket to switch its pick between a and b. Find every way to make .
Six ways — and 6 is exactly the middle number of row 4. You get one for every way of choosing which 2 of the 4 brackets give b. Counting choices is the job of
In general, the coefficient of in is the number of ways to choose the r brackets that give b:
where n is a
So every entry of Pascal's triangle can be written as a combination (Fig 7.3):
| Index | Coefficients |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 |
Now any row can be written down without the rows above it. Row 7, for instance, is , so
Tip
You never need the rows above. Counting places from 0, the number in place r of row n is . For example, the middle of row 10 is
Speed Round: Read the Triangle
Every number in Pascal's triangle is some . Work these out in your head. The triangle is symmetric, so — use whichever keeps the numbers small, and cancel before you multiply.
Get five right to move on.
7.2.1 Binomial theorem for any positive integer n
Everything you have found so far fits into one statement.
Binomial theorem. For every positive integer ,
We have checked it for small powers. To be sure it holds for every n, we prove it by the
The idea of induction. Think of a long line of dominoes. Knock over the first one, and make sure that every domino, as it falls, knocks over the
Proof. Let be the statement of the theorem.
Step 1: the first domino. For , the right-hand side is , which is . So is
Step 2: suppose one domino falls. Suppose is true for some positive integer k, that is,
Step 3: the next one falls. Write and multiply (1) once by a and once by b — exactly what the multiplying machine did:
Add the two lines and collect like terms. Each bracket takes one coefficient from each line, just like the × a and × b rows:
Each bracket is the "add the two numbers above" rule of the triangle, written with combinations. It is called Pascal's rule:
Check it on row 5:
Using Pascal's rule, together with and ,
which is exactly . So is true whenever is true. As is true, the principle of mathematical induction tells us that is true for
Put the steps of the proof in order.
That is the shape of every induction proof: a first case, and a step that carries any case on to the next one.
Term by Term
Let's put the theorem to work on . Here a = x, b = 2 and n = 6, so the terms are , , , and so on up to . The first term is done for you. Simplify the other six: tap the right answer in each row.
So
Observations
Here is what every expansion has in common. Keep these in mind and most slips get caught on their own.
1. The sigma form. The theorem can be written in one short line with the sigma sign, which means "add up the terms for k = 0, 1, 2, …, n":
where and are read as 1.
2. Binomial coefficients. The numbers that appear in the expansion are called the binomial coefficients.
3. The number of terms. has
4. The powers. The power of a starts at n and falls by 1 from term to term, down to
5. The sum of the powers. In every term the powers of a and b add up to n: , then , and so on to .
Quick check: which of these could be a term of ?
Right: the powers add up to 4 + 5 = 9, and the coefficient is .
Some special cases
(i) A minus sign inside the bracket. Put and . Odd powers of are negative and even powers are positive, so the signs of the terms alternate:
Try it on . Watch the signs as well as the numbers.
Every other term is negative: exactly the terms with an odd power of .
(ii) When the first term is 1. Put and :
Now put :
(iii) A minus sign as well. Put and :
and putting gives
So every row of Pascal's triangle adds up to a power of 2, and with alternate signs it adds up to 0.
Test both claims on the triangle.
Row
Row 8 is the one: and . There is a counting reason too: in the bracket game, every one of 8 brackets can give a or b, so there are sets of picks in all — and row 8 just sorts them by how many b's they contain.
Worked Examples
Example 1 : Expand , .
Solution : Here , and row 4 is 1, 4, 6, 4, 1. So
Simplify the three middle terms:
The first term is and the last is , so
Example 2 : Compute .
Solution : Powers of 98 are awkward, but powers of 100 and of 2 are easy. So write and expand : six easy terms instead of four hard multiplications. Because 2 is subtracted, the terms alternate in sign.
Give each term its sign and watch the total build up.
Written out in full:
Example 3 : Which is larger, or 10,000?
Make your call before you calculate.
Solution : Split 1.01 as and write only the first two terms of the expansion. All the other terms are positive:
Hence .
Tip
To compare with a number when , the first two terms are often enough: , because every term after them is positive.
Example 4 : Using binomial theorem, prove that always leaves remainder 1 when divided by 25.
A proof has to cover every n at once. Before proving the claim, test it.
Test it: change n and check the remainder each time.
Solution : When , we say that dividing a by b gives quotient q and remainder r. So we have to show that for some whole number k.
By special case (ii),
Put a =
Every term after contains the factor
where is a whole number. So leaves remainder
Speed Round: Coefficient Detective
The last challenge of the chapter: no full expansions allowed. In the term with is , so its coefficient is . Watch the sign when k is negative.
Get five right to move on.