Binomial Theorem:

We live in an Era which uses calculation as its base. Since childhood you are studying calculations based on addition, subtraction, multiplication and division. But still, you find it tough to calculate something like this (88)⁵. It’s simple to say that anyone who knows multiplication can do it but it will take much time. Yes, Of course, you can do it, also you have to save time. To care all these things there was introduced binomial theorem. It provided an easier way to expand (a + b)ⁿ, where n is an integer or a rational number. Here, we study binomial theorem for positive integral indices only.

Binomial Theorem for Positive Integral Indices:

(a+ b)⁰ = 1           a + b ≠ 0

(a+ b)¹ = a + b

(a+ b)² = a² + 2ab + b²

(a+ b)³ = a³ + 3a²b + 3ab² + b³

(a+ b)⁴ = (a + b)³(a + b) = a⁴+ 4a³b + 6a²b² + 4ab³ + b⁴

Properties of binomial Theorem based on above observation:

1. The number of terms in RHS is 1 more than the terms present in LHS.
2. In each successive terms powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the second quantity ‘b’ increase by 1.
3. In each term of expansion, the sum of powers in each individual term is same as that of original LHS of (a + b)’s power.
4. We say the coefficients ⁿC_r occurring in the binomial theorem as binomial coefficients.

Coefficient of Binomial Expansion:

Pascal’s Law made it easy to determine the coefficient of binomial expansion. Let’s see:

Suppose,
(a + b)⁵ = 1.a⁴⁺¹ + 5.a⁴b + 10.a³b² + 10.a²b³ + 5.ab⁴ + 1.b⁴⁺¹

We see here according to Pascal’s triangle that the coefficient of respective terms is same as that obtained in triangle thus you can apply the same everywhere for expansion.

• You can also obtain Pascal’s triangles n^th row by
  ⁿC₀ ⁿC₁          ⁿC₂          ⁿC₃          ⁿC_……                       ⁿC_n
• Example: For power of 3 i.e. (a + b)^{3
  3}C₀         ³C₁          ³C₂          ³C₃
   1             3              3              1
  which is same as expansion of (a + b)³
  e 1.a³ + 3.a²b + 3.ab² + 1.b³

Binomial theorem for any positive integer n:

• (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n–1b + ⁿC₂ a^n–2 b² + . … .  + ⁿC_{n – 1}b^n–1 + ⁿC_n bⁿ
  We write it as the sum in short as:
• (a + b)ⁿ = \(\sum_{K=0}^{n}\) ⁿC_k a^n-kb^k

Some Special cases for the expansion of (a + b)ⁿ :

1. When we take a = x and b = – y, then
   (x – y)ⁿ = ⁿC₀ x_n – ⁿC₁ x^{n – 1}y + ⁿC₂ x^{n – 2}y² +  …    + (–1)^{n n}C_n yⁿ
2. When we take a = 1 and b = x, then
   (1 + x) ⁿ = ⁿ C₀ + ⁿC₁ x + ⁿC₂ x² + ⁿC₃ x³ + … + ⁿC_n xⁿ.
3. When we take a = 1 and b = – x, then
   (1– x)ⁿ = ⁿ C₀ – ⁿC₁ x + ⁿC₂ x² – ⁿC₃ x³ + … + (–1)^{n n}C_n xⁿ.

General and Middle Terms of Binomial Expansion:

1. General term of binomial expansion : (r +1)^th term is said to be general term of expansion:
   ⁿC_ra^n–r.b^r
2. Middle term of binomial expansion :

• If n is even, then the number of terms in the expansion will be n + 1. Also the middle term will be \(\left( \frac{n}{2} + 1 \right)\) ^th
• If n is odd, then n +1 is even, so there will be two middle terms in the binomial expansion.
  They are: ( \(\frac{n+1}{2}\) )^th term and ( \(\frac{n+1}{2} + 1\) )^th

Binomial Theorem Examples:

1. Compute (98)⁵.
   Solution:
   to compute (98)^five. We must express 98 as the sum or difference of two numbers whose powers are easier to handle, and then we will use Binomial Theorem.
   98 = 100 – 2
   (98)⁵ = \({(100 – 2)}^{5}\)

=⁵C₀ (100)⁴⁺¹ – ⁵C₁ (100)⁴ .2 + ⁵C₂ (100)³ 2² – ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵
= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032 = 9039207968

2. Which is larger (1.01)^1000000 or 10,000?
   Solution:
   Let’s first break 1.01
   (1.01)^1000000
   = (1 + 0.01)^1000000

= ^1000000C₀ + ^1000000C₁ (0.01) + all the positive terms
= 1 + 1000000 × 0.01 + all the positive terms

= 1 + 10000 + all the positive terms > 10000
Hence,  we can say that

(1.01)^1000000 > 10000