Binomial Theorem
The binomial theorem generalizes classical formulas such as (a + b)2 using combinatorial techniques.
Motivation
Surely you know the formulas for calculating (a + b)2 and (a + b)3:
$$\begin{eqnarray} (a+b)^2&=&a^2+2ab+b^2\\ (a+b)^3&=&a^3+3a^2b+3ab^2+b^3 \end{eqnarray}$$
And a natural question is whether these formulas can be generalized for any natural number n, i.e., can (a + b)n be computed?
Derivation of
Of course it can, and the binomial theorem is used to do just that. Let's try to see how this development would proceed further for n = 4 and n = 5.
$$\begin{eqnarray} (a+b)^4 &=& a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\\ (a+b)^5&=&a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 \end{eqnarray}$$
You should already see a definite pattern. If we have the expression (a + b)n, we always start with the term an on the right-hand side. The second term also contains the variable a, but the exponent is one less. For example, in the expansion of (a + b)4, we have a4 as the first term on the right-hand side and a3 in the second term. Each subsequent term contains the variable a with an exponent that is one lower than in the previous term. On the other hand, the variable b increases in its exponent. Initially, we either do not have the variable b or it appears with an exponent of zero: b0 = 1. In the second term, b then has an exponent of one, then two, and so on.
In all this, the simple rule that the sum of the exponents must always equal n applies. If we consider the last distribution of (a + b)5, then, for example, the third term is 10a3 · b2, and the sum of the exponents of the variables a and b is precisely 5.
The last thing to discuss is what the coefficients (the numbers before the variables) will be for each term. You will notice that the coefficients are symmetric: for (a + b)4, we have the coefficients 1,4,6,4,1; for (a + b)5, we have 1, 5, 10, 10, 5, 1. I remind you that a4 = 1 · a4; this is why I include the one there, even though it's not explicitly mentioned in the formulas. Next, we will demonstrate the related pattern in Pascal's triangle.
Pascal's triangle
Pascal's triangle is a handy tool for calculations using the binomial theorem. Pascal's triangle consists of numbers arranged such that the number below any two adjacent numbers is equal to their sum. This may sound complex, but the triangle itself clarifies it:
Consider the number four, for example. Above this number are the numbers 1 and 3. And the sum of 1+3 equals exactly 4. The same is true for the other numbers.
And how do we use Pascal's triangle? If you look at the numbers on each row, you'll find that they closely resemble the coefficients in the resulting binomial expansion. For example, for (a + b)2, the coefficients are 1,2,1, which exactly match the second row. For (a + b)4, the coefficients are 1,4,6,4,1, which exactly match the fourth row. If we need to compute (a + b)n, we simply look at the n-th row of Pascal's triangle.
Definition of the binomial theorem
Formally, the binomial theorem is expressed as follows:
$$ (a+b)^n=\sum_{k=0}^n {n \choose k} a^{n-k}b^k\qquad a,b\in\mathbb{R}; n\in\mathbb{N} $$
We compute the sum from zero to n, inclusive. Thus, the resulting number of addends is n + 1. The term in parentheses is the combinatorial number, which pertains to combinatorics and represents the number of different combinations through which we can derive a specific binomial expansion term. The remainder are simply variables with corresponding exponents. Note that the sum of the exponents actually equals n: n − k + k = n.
Examples
A simple example: calculate using the binomial theorem (a + 2b)4. First, we examine Pascal's triangle on the fourth line. This reveals that the coefficients will take the form: 1,4,6,4,1. Now, all that's left is to apply these coefficients to the exponents one by one. The first term will simply be a4:
$$(a+2b)^4=a^4+\ldots$$
In the second term, the exponents are decomposed into the pairs 3,1, yielding a3(2b)1. The exponent on the first term has no effect on the expression, allowing us to write a32b. The coefficient of the second term is 4:
$$(a+2b)^4=a^4+4a^32b+\ldots$$
The third term will have exponents 2,2. Be careful here; we are multiplying the entire 2b, so this is: (2b)2, which equals: (2b)2 = 4b2. We then add to the expansion with a coefficient of 6:
$$(a+2b)^4=a^4+4a^32b+6a^24b^2+\ldots$$
The fourth term will have the form a(2b)3. After the exponentiation, we obtain a8b3. We then add it with a coefficient of 4:
$$(a+2b)^4=a^4+4a^32b+6a^24b^2+4a8b^3+\ldots$$
And the last term is simply (2b)4 = 16b4:
$$(a+2b)^4=a^4+4a^32b+6a^24b^2+4a8b^3+16b^4.$$
Now we just need to make a slight modification. In the term 4a32b, we can place a 2 at the beginning of the term and multiply by 2 · 4 = 8. This will yield 8a3b. By doing this for the others, we achieve the result:
$$(a+2b)^4=a^4+8a^3b+24a^2b^2+32ab^3+16b^4.$$
Please note that at this point, the coefficients no longer match the numbers in Pascal's triangle.