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) .