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Binomial Theorem – Concept and Applications

Thursday, August 2nd, 2018

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Binomial Theorem is a very intriguing topic in mathematics due to its wide-ranging application. Before moving forth with its use, let me ask you a question? Do you remember what’s is (a + b)² or (a-b)²? I guess most of you would have easily say (a+b)² is a² + b² + 2ab and (a² – b²) is a² + b² -2ab. But, what about (a + b)⁴? Okay, this is something you can still deduce. Let’s make it a little bit more complex, can you tell what the expansion for would be (a + b)⁷ or (7a – 5b)¹⁰. Now, I think it would not that simple to answer as you did earlier. Here’s something where the binomial Theorem can come into practice. And, in fact expansion of expressions such as is (a + b), (a-b)² or (a + b)³ have all come through the use of Binomial Theorem. I hope that now you have understood that this article is all about the application and use of Binomial Theorem.

Many of you might be well acquainted with binomial Theorem specially those who had mathematics in their high school and those of you who know nothing about you will learn everything that’s necessary about it in this article. Let’s dig a deeper meaning in the nomenclature “Binomial Theorem/ Expansion”. Now Binomial means sum or difference of two terms and Binomial Expansion is the expansion of that sum of two terms. Since we have learnt what literally the meaning is, now is the time for some mathematics.

If both x and y are real and all n N then the Binomial Expansion of

(x + y)ⁿ = ⁿC₀ xⁿ y⁰ + ⁿC₁ x^n-1 y + ⁿC₂ x^n-2 y² + ….+ ⁿC_n-1 x y^n-1 + ⁿC_n x⁰ yⁿ

                                                                                                                 = ∑ⁿ_r=0 ⁿC_r x^n-r y

Therefore,

(x + y)² = ²C₀ x² y⁰ + ²C₁ x y + ²C₂ x⁰ y²

= x² + 2xy+ y²

(x + y)³ = ³C₀ x³ y⁰ + ³C₁ x² y + ³C₂ x¹ y² + ³C₃ x⁰ y³

= x³ + 3x²y + 3xy² + y³

(x + y)⁴ = ⁴C₀ x⁴ y⁰ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴

= x⁴ + 4x³y + 6x²y²+ 4xy³ + y⁴

And so on.

Special Cases

1. (x – y)ⁿ = ⁿC₀ xⁿ y⁰ – ⁿC₁ x^n-1 y + ⁿC₂ x^n-2 y² + ….+ (-1)^{n n}C_n x⁰ yⁿ
2. (1 + x)ⁿ = ⁿC₀  y⁰ + ⁿC₁ y + ⁿC₂ y² + ….+ ⁿC_n-1 y^n-1 + ⁿC_n yⁿ
3. (1 – x)ⁿ =  ⁿC₀ – ⁿC₁ x + ⁿC₂ x² + …. +(-1)^{n n}C_n xⁿ

Tricks and Tips:

• The coefficients of (r + 1)^th term in the expansion of (1 + x)ⁿ is ⁿC_r.
• The coefficient of x^r in the expansion of (1 +x)ⁿ is ⁿC_r.

Important Points to remember:

• The positive integer n is called the index of the binomial.
• Number of terms in the expansion of (x + y)ⁿ is n + 1.
• In the expansion of (x + y)ⁿ, the power of x goes on decreasing by 1 and that of y goes on increasing by 1, therefore the sum of their powers should be equal to n.
• The binomial coefficients of the terms equidistant from the end and the beginning are equal.
• If n is odd, then the number of terms in (x + b)ⁿ + (x – b)ⁿ and in (x + b)ⁿ – (x – b)ⁿ are equal to (n + 1)/2.
• If n is even, then the number of terms in (x + b)ⁿ + (x – b)ⁿ are (n + 1)/2 and in (x + b)ⁿ – (x – b)ⁿ are n/2.
• xⁿ + yⁿ is divisible by x + y if n is odd as xⁿ + yⁿ = (x + y) (x^n^-1 – x^n^-2y + x^n^-3y² – …. + y^n^-1)
• xⁿ – yⁿ is divisible by x + y if n is odd as xⁿ + yⁿ = (x – y) (x^n^-1 + x^n^-2y + x^n^-3y² + …. + y^n^-1)

Formulas to remember

• ⁿC₀ = 1
• ⁿC₁ = n
• ⁿC₂ = n (n – 1)/2
• ⁿC₃ = n (n – 1) (n – 2)/ 3!
• ∑ⁿ_i=0 (ⁿC_i) = ²ⁿC_n
• ∑ⁿ_i=0 i (ⁿC_i)= n. 2^n-1
• ∑ⁿ_i=0 i (i – 1) … (i – k + 1) (ⁿC_i) = n (n-1)… (n – k + 1)

 

We can now move to some basic application of the above learned concept that can come in competitive exams such as CAT, XAT, IIFT, SNAP etc.

Example 1: The coefficient of x⁵ in the expansion of (1 + x²)⁵ (1 + x)⁴ is

a. 40

b. 50

c. -50

d. 60

Sol: (1 + x²)⁵ (1 + x)⁴ can be expanded and written as

(1+⁵C₁ x + ⁵C₂ x⁴ + ⁵C₃ x⁶ + ….) (1 + ⁴C₁ x + ⁴C₂ x² +…)

(1 + 5x² + 10x⁴ +…) (1 +4x + 6x² + 4x³ + x⁴)

The terms giving x⁵ in the above product is (5x²) (4x³) + (10x⁴) (4x)

⇒(20 + 40) x⁵

⇒60x⁵

Hence, the coefficient is 60.

Example 2: If 7¹⁰³ is divided by 25, then the remainder is

a. 20

b. 16

c. 18

d. 15

Sol: 7¹⁰³ = 7 (7²)⁵¹ = 7(50 – 1)⁵

= 7 (50⁵¹ – ⁵¹C₁ 50⁵⁰+ ⁵¹C₂ 50⁴⁹ – … – 1)

= 7 (50⁵¹ – ⁵¹C₁ 50⁵⁰+ ⁵¹C₂ 50⁴⁹ -…) – 7 + 18 – 18

= 7 (50⁵¹ – ⁵¹C₁ 50⁵⁰+ ⁵¹C₂ 50⁴⁹ -…) – 25 + 18

= k + 18                                                                                                     (here, k is divisible by 25)

Therefore, remainder is 18.

Example 3: Larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰is

a. 101⁵⁰

b. 99⁵⁰ + 100⁵⁰

c. Both are equal

d. None of these

Sol: 101⁵⁰ = (100 + 1)⁵⁰

⇒(100 + 1)⁵⁰ = 100⁵⁰ + 50.100⁴⁹ + [(50. 49)/ 2. 1]. 100⁴⁸ +….

and 99⁵⁰ = (100 – 1)⁵⁰ = 100⁵⁰ – 50.100⁴⁹ + [(50. 49)/ 2. 1]. 100⁴⁸ – ….

Subtracting both we get,

101⁵⁰ – 99⁵⁰ = 2(50. 100⁴⁹ + [(50. 49. 48)/ 1. 2. 3]. 100⁴⁷ + …)

= 100⁵⁰ + 2. [(50. 49. 48)/ 1. 2. 3]. 100⁴⁷ +…. > 100⁵⁰

Hence, 101⁵⁰ > 99⁵⁰ + 100⁵

Example 4: The last digit of the number (32)³² is

a. 4

b. 6

c. 8

d. None of these

(32)³² = (2 + 3. 10)³²

= 2³² + 10k, where k

Therefore, last digit of 32)³² = last digit of 2³²

But 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32

Therefore, 2³² = (2⁵)⁶. 2² = (32)⁶. 4 = (2 + 30)⁶. 4

= (2⁶ + 10r). 4, r

Hence, last digit of 32)³² = last digit of 2³² = last digit of 2⁶. 4= last digit in 4 x 4 = 6.

Few more concepts related to Binomial Expansion

Method of Finding Independent term or constant term.

i.) Write down the general term in the expansion of (x + a)ⁿe. (r + 1)^th term.

⇒ T_r+1 = ⁿC_r x^n-r a^r

ii.) Separate the constants and variables. Also, group them separately.

iii.) Since, we need them to find independent of x in the given binomial expansion, equate to zero the index of x and accordingly we’ll get the value of r for which there exists a term independent of x in the expansion.

Method of Finding Greatest Term in the expansion of (1 + x)ⁿ

Method 1

i.) Let T_r be the greatest term.

ii.) Find T_r-1, T_r, T_r+1, from the given expansion.

iii.) Put T_r/ T_r+1 ≥ 1 and T_r/ T_r-1 ≥ This will give an inequality from where value or values of r can be obtained.

iv.) Then, find the rth term T_r, which is the greatest term.

Method 2

i.) Find the value k = [(n + 1) |x|] / 1 + |x|

ii.) If k is an integer, then T_k and T_k+1 are equal and both are greatest term.

iii.) If k is not an integer, then T_{(k) + 1} is the greatest term, where (k) is the greatest integral part of k

Tricks and Tips:

• To find the greatest term in the expansion of (x + y)ⁿ, write (x + y)ⁿ = xⁿ (1 + y/x)ⁿ and then find the greatest term in (1 + y/x)ⁿ.

Middle term in the Binomial Expansion:

The middle term in the binomial expansion of (x + y)ⁿ depends upon the value of n.

i.) If n is even, then there is only one middle term i.e. (n/2 + 1) th term.

ii.) If n is odd, then there are two middle terms i.e. (n+1/ 2) th term and (n + 3)/ 2 th term.

Tricks and Tips:

• When there are two middle terms in the expansion, their binomial coefficients are equal.
• Binomial coefficient of the middle term is the greatest binomial coefficient.

Some more applications

Example 5: The greatest term (numerically) in the expansion of (2 +3x)⁹, when x = 3/2, is

a.) (5 x 3¹¹) /2

b.) (5 x 3¹³) /2

c.) (7 x 3¹³) /2

d.) None of these.

Sol: (2 +3x)⁹ = 2⁹ (1 + 3x/2)⁹ = 2⁹ (1 + 9/4)⁹

Therefore, m = | [x (n + 1)]/ (x + 1) |

= | (9/4) (9 + 1)/ (9/4) + 1| = 90/ 13 = 6 ¹²/₁₃ ≠ Integer

= The greatest term in the expansion is T_{[m] + 1} = T_{6 + 1} = T₇

Hence, the greatest term = 2⁹. T₇

= 2⁹. T_{6 + 1} = 2⁹. ⁹C₆ (9/4)⁶

= 2⁹. ⁹C₆ (9/4)⁶ = 2⁹. (9.8.7)/ 1.2.3. 3¹³/ 2¹² = (7 x 3¹³) /2

 

Example 6: If n is a positive integer, then (√3 + 1)²ⁿ – (√3 – 1)²ⁿ is

a.) An irrational number

b.) An odd positive number

c.) An even positive number

d.) A rational number other than positive integers.

Sol: (√3 + 1)²ⁿ – (√3 – 1)²ⁿ = 2[ ²ⁿC₁ (√3)^{2n – 1} + ²ⁿC₃ (√3)^{2n – 3} + ²ⁿC₅ (√3)^{2n – 5} +…]

= which is an irrational number.

I hope the above examples have given you a fair idea about such multiple usages of the Binomial Expansion. And, these are only a few sums and there can be a thousand others in which might not directly give you hint about the effectiveness of solving this expansion in problem solving. These applications can only be known and mastered with practice only. So, keep practicing!

Other posts related to Quantitative Aptitude – Modern Maths

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Permutation and Combination – Distribution of Objects
How to find Rank of a Word in Dictionary (With or Without Repetition)
Set Theory- Maximum and Minimum Values
How to solve questions based on At least n in Set Theory for CAT Exam?
Sequence and Series Problems and Concepts for CAT Exam Preparation
Basic Probability Concepts for CAT Preparation

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