Binomial Theorem Class 11 Notes Mathematics Chapter 8 - CBSE

Chapter : 8

What Are Binomial Theorem ?

Binomial

An algebraic expression is built from the constants, variables and finite number of algebraic operations i.e., addition, subtraction etc. An algebraic expression that contains only two terms is called binomial.

The binomial theorem is an algebraic formula by which we can find the value of any positive integer exponent of a binomial.

In other words, binomial theorem, helps to expand (x + y)ⁿ for any positive integral index n. The binomial theorem gives us a simple way to simplify such large terms.

History Of Binomial Theorem

Since the 4^th century BC humankind has been able to demonstrate the binomial theorem. This cube binomial was also utilised in the 5^th century AD. Halyudth, an Indian mathematician established this approach in the 10^th century AD utilishing Pascal's triangle.

In the 12^th century, this theorem was explicity established Mathematicians advanced these findings until Sir Issac Newton, in 1665, expanded the binomial theorem to all exponents.

Pascal Triangle

Pascal's triangle, in algebra, a triangular arrangements to numbers that gives the coefficients in the expansion of any binomial expression such as (x + y)ⁿ . It is named for the 17^th century French Mathematician Blaise Pascal.

Binomial Theorem For Positive Integral Index

The binomial theorem helps to expand any positive integral power of the binomial expression.

If a and b are real numbers, then for all n ∈ N.

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n–1b + ⁿC₂ a^n–2b² +...+ ⁿC_n–1 ab^n–1 + ⁿC_nbⁿ

Proof of Binomial Theorem

Proof: This theorem is proved by using the principle of mathematical induction.

Let P(n) be the statement

(a + b)ⁿ= ⁿC₀ aⁿ + ⁿ C₁ a^n–1 b + ⁿC₂a^n–2b² + ... + ⁿC_n–1 ab^n–1 + ⁿC_n bⁿ

Then, the statement P(1) is :

(a + b)¹ = ¹C₀ a¹ + ¹C₁ a⁰ b = (a + b), which is true.

Let P(m) be true. Then,

(a + b)^m = ^mC₀ a^m + ^mC₁ a_m–1 b + ^mC₂ a^m–2 b² + ... + ^mC_m–1 ab^m–1 + ^mC_mb^m ...(i)

Multiply both sides of (i) by (a + b), we get

(a + b)^m+1 = ^mC₀ a^m+1 + ^mC₀a^mb + ^mC₁a^mb + ^mC₁ a^m–1b² + ^mC₂ a^m–1b² + ^mC₂a^m–2b³ +...+ ^mC_m−1 a² b^m−1 + ^mC_m–1ab^m+ ^mC_mab_m + ^mC_mb^{m + 1}

= ^mC₀ a^m+1 + (^mC₀ + ^mC₁)a^mb + (^mC₁ + ^mC₂)a^m–1b² +... + (^mC_m–1 + ^mC_m)ab_m + ^mC_mb^m+1

= ^m+1C₀ a^m+1 + ^m+1C₁ a^mb + ^{m + 1}C₂ a^m−1b² + ... + ^m+1C_mab^m + ^m+1C_m+1 b^m+1

$$\begin{bmatrix}\therefore\space ^{m}\text{C}_{0} = 1 = ^{m+1}\text{C}_{0},\\^{m}\text{C}_{m}1= 1 =^{m+1}\text{C}_{m+1}\\\text{and}\space ^{m}\text{C}_{r-1} + ^{m}\text{C}_{r} = \space^{m+1}\text{C}_{r}\end{bmatrix}$$

This shows that P(m + 1) is true, whenever P(m) is true. Hence, by the principle of mathematical induction the theorem is true for all n ∈ N. So, the above result can be written as,

$$(a+b)^{n} =\sum^{n}_{r =0}\text{C}_{r}a^{n-r}b^{r}$$

Binomial Theorem By Positive Integral Indices

Observe the following identities :

(a + b)⁰ = 1
(a + b)¹ = a + b
(a + b)² = a² + 2ab + b²
(a + b)³ = a³ + 3a² b + 3ab² + b³
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

From this we conclude that

The exponent of a in the expansion of (a + b)ⁿ is decreasing by 1 from n to 0.
The exponent of b in the expansion of (a + b)ⁿ increasing by 1 from 0 to n.
The sum of the exponents of a and b in the expansion of (a + b)ⁿ is always n.
The number of terms in the expansion of (a + b)ⁿ is always n + 1.
The coefficients of terms in above expression are actually making a Pascal's triangle.

Similarly, (a + b)⁵ , (a + b)⁶ can be expanded using the Pascal's triangle. But working with Pascal's triangle can be very time consuming if the exponent is a very large number (say 100).

That is why, binomial theorem is used for the expansion of (x + y)ⁿ for any positive integral index.