Applications Of Binomial Theorem

What is a binomial expression?

An algebraic expression with two terms is called a binomial -x+y or a-b. 

What is a binomial expansion or binomial theorem?

When a binomial term is raised to a non-zero positive exponent (except 1), the binomial is expanded. But when the exponents are bigger numbers, it is a tedious process to find the solution manually. Here, binomial expansion formulas are used. 

Formula of binomial theorem:

Let n ∈ N,x,y,∈ R then

(x + y)n= nΣr=0 nCr xn-r . yr 

Important points regarding binomial expansion: 

• The total number of terms in the expansion of (a+b)n is (n+1).

• The sum of exponents of a and b is always n.

• nC0, nC1, nC2, … .., nCn are binomial coefficients and can also be written as C0, C1, C2, ….., Cn.

• The equidistant binomial coefficients from the beginning and from the ending are equal; nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,….. etc.

• Binomial coefficients can also be found using Pascal’s Triangle.

Applications of binomial theorem

The binomial theorem is used in various fields of mathematics and statistics. From finding a remainder to finding a digit of a number. The most common binomial theorem applications are:

Finding remainder 

Example:

 What remainder we get when 7102 is divided by 25

Answer: 

(7102 / 25) = [(49)51 / 25)]

= [(50 − 1)51/ 25]

= [(25L − 1) / 25] 

= [(25L+24) / 25]

Therefore  The remainder =24

Finding digits of a number

Example: 

What are the last two digit  of the number (11)10

Answer: 

(11)10= (10+1)10 

= 10C0 (10)10 +10C1 (10)9 + 10C2(10)8+ 10C3 (10)7 + 10C4(10)6 +10C5(10)5…..

After 10C1(10)+10C0 all the terms will have last two digits as 0.

∴ The last two digits are 01.

Relation between two numbers

Example: which one be larger of 9925 + 10025and 10125

Answer:  9925= (100 − 1)25= 10025 – 25 . 10024+ 25 . 12 . 10023− ….

⇒10125= (100 + 1)25= 10025 + 25 . 10024 + 25 . 12 . 10023+ …

⇒ 10150– 9950= 2[25 . 10024 + 25(24) (16) 10022+ …]

= 10025+ 50 . 49 . 16 . 10047+ … >10050

∴ 10125– 9925> 10025

⇒ 10125> 10025 + 9925

Divisibility test

Example:

 Show whether 129+ 89 is divisible by 10.

Answer: 

129 + 811 = (10 + 2)9+ (10 − 2)9

= (9C0 . 109 + 9C1 . 108 .21+ … 9C9) + (9C0 . 109 − 9C1 . 10921+ … −9C9.29)

= 9C0 . 109 + 9C1 . 108 .21+ … + 9C8 . 10.28+ 29 + 109 − 9C1. 108.21+ … + 9C8 . 10.28−29

= 10[9C0 . 108 + 9C1. 10721+ … + 9C8 .28+ 108− 9C1. 107.21 + … + 9C8.28]

= 10K, which is multiple of 10.

Formulae:

• The number of terms in the expansion of (x1 + x2 + … xr)n is (n + r − 1)Cr-1

• Sum of the coefficients of (ax + by)n is (a + b)n

Binomial theorem formula and Binomial theorem calculator for any index:

If n is a rational number and x is a real number such that | x | < 1, then

Binomial theorem for negative index

If rational number and -1 < x <1 then,

• (1 − x)-1= 1 + x + x2+ x3 + … + xr+ … ∞

• (1 + x)-1 = 1 – x + x2 – x3 + … (−1)r xr+ … ∞

• (1 − x)-2 = 1 + 2x + 3×2 − 4×3 + … + (r + 1)xr+ … ∞

•  (1 + x)-2 = 1 − 2x + 3×2 − 4×3 + … + (−1)r (r + 1)xr+ … ∞

Some other valuable expansions:

(x + y)n+ (x−y)n = 2[C0 xn + C2 xn-2 y2 + C4 xn-4 y4 + …]

(x + y)n – (x−y)n = 2[C1 xn-1 y + C3 xn-3 y3 + C5 xn-5 y5 + …]

(1 + x)n = [C0 + C1 x + C2 x2 + … Cn xn]

(1+x)n+ (1 − x)n = 2[C0 + C2 x2+C4 x4 + …]

(1+x)n − (1−x)n = 2[C1 x + C3 x3 + C5 x5 + …]

The number of terms in the expansion of (x + a)n + (x−a)n are (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.

The number of terms in the expansion of (x + a)n − (x−a)n is (n/2) if “n” is even or (n+1)/2 if “n” is odd.

Question 1: Find the positive value of λ for which the coefficient of x3 in the expression x3[√x + (λ/x2)]10 is 720.

Solution:

⇒ x3 [10Cr . (√x)10-r . (λ/x2)r] = x3 [10Cr . λr . x(10-r)/2 . x-2r]

= x3 [10Cr . λr . x(10-5r)/2]

Therefore, r = 2

Hence, 10C2 . λ2 = 720

⇒ λ2 = 16

⇒ λ = ±4.

Question 2: What is the sum of the real values of x for which the middle term in the binomial expansion of (x3/2 + 2/x)8 equals 5670?

Solution:

T5 = 8C4 × (x12/16) × (16/x4) = 5670

⇒ 70 x8 = 5670

⇒ x = ± √3.

Multinomial expansions

Let us expand (a+ b + c)10. When we expand this expression, each term has different powers of a, b, c and the sum of these powers is always 10.

For example, in the term λa2b3c5, the coefficient of this term is equal to the number of ways 2a′s, 3b′s, and 5c′s are arranged, i.e., 10! (2! 3! 5!).