Binomial Theorem

Binomial expression is an algebraic expression with two terms only, e.g. 4x²+9. When such terms are needed to expand to any large power or index say n, then it requires a method to solve it.

Therefore, a theorem called Binomial Theorem is introduced which is an efficient way to expand or to multiply a binomial expression. Binomial Theorem is defined as the formula using which any power of a binomial expression can be expanded in the form of a series.

Binomial theorem

Binomial Theorem is used to solve binomial expressions in a simple way. It gives an expression to calculate the expansion of (a+b)ⁿ for any positive integer n. The Binomial theorem is stated as:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1 b¹ + ⁿC₂ a^n-2 b² + …. + ⁿC_r a^n-r b^r + …. + ⁿC_n bⁿ

Now, as it is understood that what is a binomial expression and the purpose of the binomial theorem, so try to expand (a+b)ⁿ for large values of n (e.g. n = 10. 11, 12,…) using the above statement as:

• (a + b)¹⁰ = a¹⁰ + 10a⁹b + 45a⁸b² + 120a⁷b³ + 210a⁶b⁴ + 252a⁵b⁵ + 210a⁴b⁶+120a³b⁷+ 45a²b⁸+ 10ab⁹+ b¹⁰
• (a + b)¹¹ = a¹¹ + 11a¹⁰b + 55a⁹b² + 165a⁸b³ + 330a⁷b⁴ + 462a⁶b⁵ + 462a⁵b⁶ + 330a⁴b⁷ + 165a³b⁸ + 55a²b⁹ + 11ab¹⁰ + b¹¹
• (a + b)¹² = a¹² + 12a¹¹b + 66a¹⁰b² + 220a⁹b³ + 495a⁸b⁴ + 792a⁷b⁵ + 924a⁶b⁶ + 792a⁵b⁷ + 495a⁴b⁸ + 220a³b⁹ + 66a²b¹⁰ + 12ab¹¹ + b¹²

and so on.

In case of the expansion of smaller powers, it is difficult to calculate the coefficients of the binomial expansion using the same statement. This drawback of the binomial theorem is resolved by Pascal’s Triangle.

The Pascal Triangle is an alternative method of the calculation of coefficients that come in binomial expansions, using a diagram rather than algebraic methods.

In the diagram shown below, it is noticed that each number in the triangle is the sum of the two directly above it. This pattern continues indefinitely to obtain coefficients of any index of the binomial expression.

When we observe the pattern of the coefficients of the expansion (a + b)ⁿ, The Pascal’s triangle for the pattern of the coefficients of the expansion (a + b)⁷ is shown in the figure below:

Thus, from the above diagram the expansion of small powers of n (e.g. n=0, 1, 2, 3, 4, 5, 6, 7) can be calculated as:

• (a + b)⁰ = 1
• (a + b)¹ = a + b
• (a + b)² = a² + 2ab +b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
• (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
• (a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶
• (a + b)⁷ = a⁷ + 7a⁶b + 21a⁵b² + 35a⁴b³ + 35a³b⁴ + 21a²b⁵ + 7ab⁶ + b⁷

In this way the expansion from (a + b)⁰ to (a + b)⁷ are obtained by the application of Pascal’s Triangle. But finding (a + b)¹⁵ is really a long process using Pascal’s triangle. So, here is where the binomial theorem comes into the picture.

When the coefficients are observed closely theorem become as follows:

(a+b)¹ = ¹C₀a + ¹C₁b

(a+b)² = ²C₀a² + ²C₁ab + ²C₂b²

(a+b)³ = ³C₀a³ + ³C₁a²b + ³C₂ab² + ³C₃b³

So from the above pattern, the expansion of (a + b)ⁿ becomes as follows:

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b¹ + ⁿC₂a^n-2b² + …. + ⁿC_ra^n-rb^r + …. + ⁿC_nbⁿ

The above statement is called Binomial Theorem.

Binomial Theorem Formula

The formula obtained by the Binomial Theorem is called the Binomial Theorem Formula, this formula can directly applied to a binomial equation (let it contains terms as x and y) raised to any power n is given as:

Some examples on the above formula are as follows:

• (x+y)²=x²+2xy+y²
• (x+y)³=x³+3x²y+3xy²+y³

Properties of the Binomial Theorem

The Binomial expansion or Binomial theorem is an important part of Arithmetic Mean. It is neither too easy nor too tough because it can be directly applied in a question. Some of the important properties of a Binomial expansion (a+b)ⁿ are stated as:

1. There are (n+1) terms in an expansion of a binomial expression with index n i.e. one more than the power (index) of the binomial expression. e.g, The number of terms presents in the expansion of (a+b)ⁿ is equal to (n+1).
2. The expansion of (a+b)ⁿ has first term is equal to aⁿ while the last one is equal to bⁿ.
3. Such that the sum of the indices of each a and b is equal to n only.
4. When a = b or a+b = n only, then ⁿC_a = ⁿC_b.
5. The coefficient of each term equidistant from the beginning and the end are equal. Such coefficients are called as the binomial coefficients and ⁿC_r = ⁿC_n-r where r is 0,1,2,…,n.

Sample Problems on Binomial Theorem

Let’s try some sample problems on the binomial theorem.

Problem 1: Expand the binomial expression (2x + 3y)².

Solution:

(2x)² + 2(2x)(3y) + (3y)²

= 4x² + 12xy + 9y²

Problem 2: Expand the following (1 – x + x²)⁴

Solution:

Put 1 – x = y.

Then,

(1 – x + x²)⁴ = (y + x²)⁴

= ⁴C₀y⁴(x²)⁰ + ⁴C₁y³(x²)¹ + ⁴C₂y²(x²)² +⁴C₃y(x²)³ +⁴C₄(x²)⁴

= y⁴ + 4y³x² + 6y²x⁴ + 4yx⁶ + x⁸

= (1 – x)⁴ + 4(1 – x)³x² + 6(1 – x)²x⁴ + 4(1 – x)x⁶ + x⁸

= 1 – 4x + 10x² – 16x³ + 19x⁴ – 16x⁵ + 10x⁶– 4x⁷ + x⁸

Problem 3: Find the 4th term from the end in the expansion of ((x³/2) – (2/x²))⁸.

Solution:

Since r^th term from the end in the expansion of (a + b)ⁿ is

(n – r + 2)^th term from the beginning.

Therefore, 4^th term from the end is 8 – 4 + 2,

i.e., 6th term from the beginning, which is given by

T₆ = ⁸C₅(x³/2)³(-2/x²)⁵

= ⁸C₃(x⁹/8)(-32/x¹⁰)

= -224/x

Problem 4: Find the middle term (terms) in the expansion of ((p/x) + (x/p))⁹

Solution:

Since the power of binomial is odd. Therefore, we have two middle terms

which are 5th and 6th terms. These are given by

T₅ = ⁹C₄(p/x)⁵(x/p)⁴

= ⁹C₄(p/x)

= 126(p/x)

T₆ = ⁹C₅(p/x)⁴(x/p)⁵

= ⁹C₅(x/p)

= 126(x/p)