A binomial can be viewed as an algebraic expression denoting the sum of two unlike terms. Examples of binomials are x + y, 3a -2b, x2 + y2 and so on. When the powers of binomials are expanded, the resulting polynomials display certain pattern. Binomial theorem summarizes the pattern in binomial expansions.

Binomial expansions and coefficients finds application in many other mathematical studies like binomial probabilities. Even though the general binomial expansion is intimidating in the first look, once the pattern is recognized, it is a very interesting mathematical concept to work with.

Binomial expansions are the polynomials resulting from expanding powers of binomials. The binomials are expanded by repeatedly multiplying the binomials as required.

The binomial expansions for powers 0 to 5 are given below:


Pascal's triangle is the pattern got by picking the coefficients from successive binomial expansions and arranging them in the same order.

Pascal Binomial Expansion

The following pattern can be observed from Pascal's triangle. A number in a row is got by adding the two consecutive numbers in the row above.
Generalizing the pattern in binomial expansions we can write the expansion for the nth power as,
(x + y)n = P0xn + P1xn-1y + P2xn-2y2 +........... + Pn-1x2yn-1 + Pnyn, where the coefficients P0, P1, ....., Pn are numbers in the n row of the Pascal's triangle.

Binomial theorem states the binomial expansion for a positive integer as follows:

(x + y)n = $x^{n}+nx^{n-1}y$ + $\frac{n(n-1)}{1.2}$$x^{n-2}y^{2}$ + $\frac{n(n-1)(n-1)}{1.2.3}$$x^{n-3}y^{3}+........+y^{n}$

Binomial expansion theorem can also be stated replacing the coefficients with combination notation as follows:
(x + y)n = $^nC_{0}x^{n}+^nC_{1}x^{n-1}y+^nC_{2}x^{n-2}y^{2}+^nC_{3}x^{n-3}y^{3}+...........+^nC_{n}y^{n}$

The Binomial theorem can be proved using the technique of mathematical induction.

The patterns in the Binomial expansion can be listed as follows:
  1. The expansion of (x + y)n has (n + 1) terms.
  2. The first term is xn and the last term is yn.
  3. The Power of x decreases by 1 in successive terms, while the power of y increases by the same number 1.
  4. Each term in the expansion is of degree n.
  5. The coefficients are symmetric. That is the rth term and (n - r)th term have the same coefficient.
  6. The Binomial expansion has one middle term the $($$\frac{n}{2}$$+1)^{th}$ if n is even, and has two middle terms, the $($$\frac{n+1}{2}$$)^{th}$ and $($$\frac{n+1}{2}$$+1)^{th}$ terms when n is odd.
The binomial expansion can be written in condensed form using Sigma and factorial notations as follows:
(x + y)n = $\sum_{r=0}^{n}$$\frac{n!}{r!(n-r)!}$$x^{n-r}y^{r}$

The general term or the (r +1)th term in the expansion is given by
Tr+1 = $\frac{n!}{r!(n-r)!}$$x^{n-r}y^{r}$
Find the 4th term in the expansion of (2a + 3)6.
Here x = 2a, y = 3, n = 6 and r + 1 = 4 hence r = 3

Using the general term formula

T4 = $\frac{6!}{3!3!}$$(2a)^{3}(3)^{3}$ = 20(8a3)(27)

Simplifying the above expression we get the 4th term in the expansion as 4320a3 .
Negative Binomial Expansion gives place to an infinite series as follows with a condition.

(1 + x)n = $1$+$\frac{nx}{1!}$+$\frac{n(n-1)}{2!}$$x^{2}+$$\frac{n(n-1)(n-2)}{3!}$$x^{3}+..........$ where |x| < 1.

In fact the above expansion is valid for all real values of n. When n is a positive integer, this expansion gives the same expansion as Binomial theorem with a finite n + 1 terms.

(1 + x)-4 = $1+(-4)x+$$\frac{(-4)(-5)}{2!}$$x^{2}+$$\frac{(-4)(-5)(-6)}{3!}$$x^{3}+..........$
= 1 - 4x + 10x2 - 20x3 + ........

The coefficients in negative binomial expansion are alternatively positive and negative.

Solved Examples

Question 1: Use Pascal's triangle to expand (2x - y)7.
Using the notations of binomial expansion x = 2x, y = -y and n = 7
    Hence the 7th row in Pascal's triangle gives the coefficients for the binomial expansion.
    7th row elements of the Pascal's triangle = 1, 7, 21, 35, 35, 21, 7, 1
    Thus with powers of 2x decreasing and the powers of -y increasing, the expansion can be written as
    (2x - y)7 = 1.(2x)7 + 7(2x)6(-y) + 21(2x)5(-y)2 + 35(2x)4(-y)3 + 35(2x)3(-y)4 + 21(2x)2(-y)5 + 7(2x)(-y)6 + 1(-y)7.
              = 128x7 - 448x6y + 672x5y2 - 560x4y3 + 280x3y4 - 84x2y5 + 14xy6 - y7.

Question 2: One of the terms contain a3 in the expansion of (4a + 3b)7.
    1.  What is the coefficient in this term?
    2. Find the exponent of b in the term.
This problem is solved considering the formula for the general term in a Binomial Expansion.

    Tr + 1 = $\frac{n!}{r!(n-r)!}$$x^{n-r}y^{r}$

    Rewriting this for the above expansion with x = 4a, y = 3b and n = 7

    Tr + 1 = $\frac{7!}{(7-r)!r!}$$(4a)^{7-r}(3b)^{r}$

    It is given that one of the terms contains a3.  Hence 7 - r = 3  ⇒    r = 4.

    Thus the 5th term contains a.

    T5 = 35(4a)3(3b)4  = 181440a3b4.

    The coefficient in this term = 181440.

    The power of b in this term = 4.