Formula sheet: Binomial Theorem
Binomial Theorem16 min readLesson 1 of 5
1.1 The expansion
For a positive integer n, the binomial theorem expands the nth power of a sum into n plus 1 terms. The powers of a fall from n to 0 while the powers of b rise from 0 to n, and the sum of the two powers is n in every term. The coefficient of the term with r factors of b is nCr, so the coefficients are the binomial coefficients of the nth row.
The signs depend on b. When b is negative, the terms alternate in sign, which is the standard way to expand a difference. Pascal's triangle gives the same coefficients row by row, and the identity nCr plus nC(r minus 1) equals (n plus 1)Cr is the rule that builds each row from the previous one.
| Expansion | Coefficients |
|---|---|
| (a plus b) to the n | nC0, nC1, nC2, ..., nCn |
| (a minus b) to the n | alternating signs on the same coefficients |
| (1 plus x) to the n | nC0, nC1, nC2, ..., nCn |
| (1 minus x) to the n | alternating signs |
1.2 General term and middle terms
The general term of the expansion of (a plus b) to the n is the (r plus 1)th term, written T(r plus 1) equal to nCr times a to the (n minus r) times b to the r. This one formula answers every question about a particular power of x, because you set the total power of x in the general term equal to the required value and solve for r.
If r comes out as a whole number between 0 and n, the term exists; otherwise some power of x is missing and the coefficient is zero. Once r is fixed, substitute it in the general term to get the coefficient or the term itself.
The middle term depends on the parity of n. If n is even, there is one middle term at position n by 2 plus 1. If n is odd, there are two middle terms at positions (n plus 1) by 2 and (n plus 3) by 2. For n equal to 10 the single middle term is the 6th, and for n equal to 9 the two middle terms are the 5th and the 6th.
1.3 Coefficient sums worth memorising
Setting x equal to 1 in (1 plus x) to the n gives the sum of all binomial coefficients as 2 to the n. Setting x equal to 1 in (1 minus x) to the n gives the alternating sum as 0, which says that the odd and even coefficient sums are equal. Those two facts solve most questions that ask for a sum of coefficients.
| Substitution | Result |
|---|---|
| x = 1 in (1 plus x) to the n | nC0 plus nC1 up to nCn = 2 to the n |
| x = minus 1 in (1 plus x) to the n | alternating sum = 0 |
| x = 1 with doubled coefficients | even-position sum = odd-position sum |
1.4 Approximations and divisibility
For a small value of b, the first two or three terms of (1 plus b) to the n give a good approximation, and the terms that are dropped are positive, so the truncated expansion is a lower bound when b is positive. This is the standard way to compare a large power with a number without computing it exactly. For example, (1.01) to the million starts 1 plus 10000 plus positive terms, so it is larger than 10000.
Divisibility problems use the same idea in reverse. Write the base as 1 plus a convenient number, expand, and collect the terms that contain the divisibility modulus. To show that 6 to the n minus 5n leaves remainder 1 on division by 25, expand (1 plus 5) to the n and note that every term from the third onward carries a factor of 25.
Exam
For a term question, write the general term first and simplify its power of x before doing anything else. For a coefficient-sum question, substitute x equal to 1 or minus 1 instead of expanding.