# Binomial Theorem – Concept and Applications

Thursday, July 16th, 2020 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 xn-1 y + nC2 xn-2 y2 + ….+ nCn-1 x yn-1 + nCn x0 yn

= nr=0 nCr xn-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 y0nC1 xn-1 y + nC2 xn-2 y2 + ….+ (-1)n nCn x0 yn
2. (1 + x)n = nC0  y0 + nC1 y + nC2 y2 + ….+ nCn-1 yn-1 + nCn yn
3. (1 – x)n =  nC0nC1 x + nC2 x2 + …. +(-1)n nCn xn

Tricks and Tips:

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

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!
• ni=0 (nCi) = 2nCn
• ni=0 i (nCi)= n. 2n-1
• ni=0 i (i – 1) … (i – k + 1) (nCi) = 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 (505151C1 5050+ 51C2 5049 – … – 1)

= 7 (505151C1 5050+ 51C2 5049 -…) – 7 + 18 – 18

= 7 (505151C1 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.

⇒ Tr+1 = nCr xn-r ar

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 Tr be the greatest term.

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

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

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

Method 2

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

ii.) If k is an integer, then Tk and Tk+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 = T6 + 1 = T7

Hence, the greatest term = 29. T7

= 29. T6 + 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!

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### 2 responses to “Binomial Theorem – Concept and Applications”

1. PSB says:

Great!

2. Sanjeev Kumar says:

Thanks

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