Introduction to Binomial Theorem

A Binomial Expression

Any algebraic expression consisting of only two terms is known as a binomial expression. It's expansion in power of x is shown as the binomial expansion.

For example: (i)  a + x   (ii)   a² + 1/x²  (iii)  4x - 6y

Binomial Theorem

Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.

It can be easily understood by examples.

        (a + x)² = a² + 2ax + x²

        (a + x)² = a³ + 3a²x + 3ax² + x³

Here, we see that the expression of (a + x)² is simple, we just multiply (a + x) by (a + x). Expansion of (a + x)³ is little tougher, but what happens when the expansion is raised to the power of ten or more? So, we have to establish the formula for (a + x)ⁿ, where n is any integer. Let us define 'a' as the first term, 'x' as the second term and 'n' as the exponent.

The total terms in the expansion of (a + x)² and (a + x)³ are 3 and 4 respectively, which means that the number of terms in the expansion is one more than the exponent. So total number of terms in the expansion (a + x)ⁿ is (n + 1).

        Now, for n = 2

           (a + x)² = a²x⁰ +  2a¹x¹/1 + (2(2-1)/1*2)a⁰x²

        = (F.T.)ⁿ (S.T.)⁰ + n/1! (F.T.)^(n-1) (S.T.)¹ + (n(n-1)/2!) (F.T.)^(n-2) (S.T.)²

        = a² + 2ax + x²

Note: F.T. refers to first term i.e. 'a' and S.T. refers to second term i.e. 'x' 

Similarly,

        (a + x)³ = a³ + 3a²x + ((3(3-1)/(1*2))a⁰ x² + ((3(3-1)(3-2))/(1*2*3))x³

                     = a³ + 3a²x + 3ax² + x³

When n is a positive integer, then

(a + x)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1x + ⁿC₂a^n-2 x² +...+ ⁿC_ra^n-r x^r +...+ ⁿC_nxⁿ,

Where ⁿC₀ . ⁿC₁ . ⁿC₂ ... ⁿC_n are called Binomial coefficients.
 

Proof of Binomial Theorem

Proof of Binomial Theorem is very simple; we can prove it by using the mathematical induction.

Proof:

Step I:      Let n = 1

                L.H.S. = a + x

                R.H.S. = a + ¹C₁ x = a + x

        So, theorem is true for n = 1.

Step II:

        Let the theorem be true for n = m, than

        P(m) : (a + x)^m = ^mC₀ a^m + ^mC₁ a^m-1 x¹ + ^mC₂ a^m-2 +...+ ^mC_m x^m       (i)

Step III:

        We have to prove for n = m + 1 i.e. we have to prove that

P(m + 1): (a + x)^m+1 = ^m+1C₀ a^m+1 + ^m+1C₁ a^m x¹

+...+ ^m+1C_m a x^m + ^m+1C_m+1 x^m+1             ....(ii)

        Multiplying by (a + x) on both sides in equation (i), we get,

        (a + x)^m+1 = (^mC₀ a^m+1 + ^mC₁ a^m +...+ ^mC_m-1 a² x^m-1 + ^mC_m a x^m)

                        = (^mC₀ a^m x + ^mC₁ ^mC₀) a^m x +...+ ^mC_m-1 a x^m + ^mC_m x^m+1)

Or, (a+x)^m+1 = ^mC₀ a^m+1 + (^mC₁ + ^mC₀) a^m x +...+ (^mC_m-1 + ^mC_m-2) a² x^m-1

+ (^mC_m + ^mC_m-1) a x^m + ^mC_m x^m+1

                        = ^m+1C₀ a^m+1 + ^m+1C₁ a^m x + ^m+1C₂ a^m-1 x² +...+

^m+1C_m-1 a² x^m-1 + ^m+1C_m a x^m + ^m+1C_m+1 x^m+1.      

Hence, Proved.

Illustration:

        Expand  (x + 1/x)⁷ .

Solution:

         (x + 1/x)7 = ⁷C₀x⁷ + ⁷C₁x⁶  (1/x) + ⁷C₂x⁵ (1/x²) + ⁷C₃x⁴ + 1/x³ + ⁷C₄x⁴

(1/x³)  + ⁷C₅x² + (1/x⁵) + ⁷C₆x (1/x⁶) + ⁷C₇ (1/x⁷)

        = x⁷ + 7x⁵ + 21x³ + (35 x) + (35/x )+( 21/x³) + (7/x⁵)+ (1/x⁷) .
 

IIT JEE study material is available online free of cost at askIITians.com. Study Physics, Chemistry and Mathematics at askIITians website and be a winner. We offer numerous live online courses as well for live online IIT JEE preparation - you do not need to travel anywhere any longer - just sit at your home and study for IIT JEE live online with askIITians.com.

To read more, Buy study materials of Binomial Theorem comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.