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Binomial Theorem for Positive Integral Indices states that “the total number of terms in the expansion is one more than the index”. It is used in the expansion of a+bn, where n is a rational number. In the binomial theorem, the powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the second quantity ‘b’ increase by 1, in successive terms. The above statement can be explained by the following identities: a + b0 = 1, where a + b0, a + b1 = a + b. Thus,
| (a + b)2 = a2 + 2ab +b2, (a + b)3 = a3 + 3a2b + 3ab2 + b3 |
- Binomial Theorem is a process of expanding an expression which has been raised to a finite power.
- As per the concept of elementary algebra, the binomial theorem (or binomial expansion) helps to describe the algebraic expansion of powers of a binomial.
- The mathematical expression which contains two terms including addition or subtraction operations is known as the Binomial Expression.
Read Also: Standard Identities
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Key Terms: Binomial Theorem, Pascal’s Triangle, Binomial Expansions, Exponent, Positive Integer Indices, Binomial Expression, Variables, Complex Numbers
What is Binomial Theorem?
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The binomial theorem represents the principle for expanding the given expression (x + y)n, thus expressing it as a sum of the terms consisting individual exponents of variables x and y. All the terms in a binomial expansion is known to be associated with a numeric value, also known as, the coefficient.
As per the Binomial Theorem Statement:
| “For any positive integer n, the nth power of the sum of two numbers a and b can also be represented as the sum of n + 1 terms of the form.” |
The expression of the Binomial Theorem Formula can be denoted by:
→ \((x+y)^n \sum_{k=0}^{n} {n \choose k} x^{n – k} y^k \)
Solved ExampleQues. What is the larger value of 9950 + 10050 and 10150? Ans. As per the given question, it can be represented as: 10150 = (100 + 1)50 = 10050 + 50 . 10049 + 25.49 . 10048 + … ⇒ 9950 = (100 − 1)50 = 10050 – 50 . 10049 + 25.49 . 10048 − …. ⇒ 10150 – 9950 = 2[50 . 10049 + 25(49) (16) 10047 + …] = 10050 + 50 . 49 . 16 . 10047 + … >10050 ∴ 10150 – 9950 > 10050 ⇒ 10150 > 10050 + 9950 |
Proof of Binomial Theorem
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Using mathematical induction, we can say,
P(n): (a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ......… nCn-1 a bn-1 + nCn bn
For n equal to 1,
P(1): (a+ b)1=1C0 a1+1C1 b1 = a+b, which is true.
Let us suppose that P(k) is true for any positive integer k.
Then, we get,
P(k): (a+b)k = kC0ak+kC1ak-1b + kC2ak-2b2+ ......… kCk-1abk-1+kCkbk............... Equation (1)
⇒ P(k+1): (a+b)k+1 = (k+1)C0ak+1+(k+1)C1akb+(k+1)C2ak-1b2 +......… + (k+1)Ck+1bk+1
⇒ (a+b)k+1
⇒ (a+b)k+1=(a+b) (a+b)k
⇒ (a+b) (kC0ak+kC1ak-1b+kC2ak-2b2 + ......… kCk-1abk-1+kCkbk)
⇒ kC0ak+1 + kC1akb + kC2ak-1b2 + ….. kCk-1a2bk-1 + kCkabk + kC0akb + kC1ak-1b2 + kC2ak-2b3+ ......... kCk-1abk + kCkbk+1
Grouping the like terms, we get
⇒ kC0ak+1+(kC1+kC0)akb+(kC2+kC1 )ak-1b2+.................. (kC1 + kCk-1 ) abk+kCkbk+1
⇒ kC0 =1=k+1C0; kCr+kCr-1=k+1Cr, and kCk=k+1 Ck+1=1
⇒ (a+b)k+1= k+1C0ak+1+(k+1)C1akb+(k+1)C2ak-1b2+.................. (k+1)Ckabk+(k+1)Ck+1bk+1
P (k+1) is true whenever P(k) is true.
Therefore, P(n) is true for all positive integral values of n.
Binomial Theorem and Pascal Triangle
Read Also: Factorization of Polynomials
Binomial Theorem for Positive Integral Indices Statement
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As per the Binomial theorem, “the total number of terms in an expansion is always one more than the index of the expansion.”
According to the expansion of (a + b)n (the number of terms here is n+1), while the index of the expression (a + b)n is n (n, here, is any positive integer greater than zero).
To expand (x + y)n, Binomial theorem can be used, where n is any rational number. Here, the expansion of (x+y)n, wherein n is an integer:
(m+n)0 = 1
⇒ (m+n)¹ = (m+n)
⇒ (m+n)² = m² + 2mn + n²
⇒ (m+n)³ = m³ + 3m²n + mn² + n³
⇒ (m+n)4 = m4 + 4 m³n + 6m²n² + 4mn³ +n4
What is Binomial Theorem for Positive Integral Indices?
Binomial Theorem in case of an index as positive integer n is represented as:
| (a + b)n = nC0an + nC1an-1b + nC1an-1b2 + … + nCn-1abn-1 + nCnbn |
Since we are aware that nCr = \(\frac{n!}{r!\ (n\ -\ r)!}\), wherein n = a non-negative integer, while 0 ≤ r ≤ n. Here, note that nCn and nC0 are equivalent to 1.
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What is Pascal’s Triangle?
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Pascal’s Triangle can be defined as the triangular arrangement of numbers which gives the coefficients in the expansion of a binomial expression. The formula to find the entry of an element in the nth row and kth column of a given pascal’s triangle, which is, \({n \choose k}\).

Pascal's Triangle
Thus, the formula of Pascal’s Triangle is:
| \({n \choose k}= {n-1 \choose k-1}+ {n-1 \choose k}\) |
Example of Pascal's Triangle
Example: What are the coefficients of expansions of (x+y)2 after using Pascal’s triangle?
Ans: Since it is clear that the coefficients of expansion of (x+y)2 are the elements in the second row of Pascal’s triangle.
Thus, the elements of the 2nd row of Pascal’s triangle are 1, 2, 1.
Hence, the coefficients of the expansion of (x+y)2 are 1, 2, 1.
Read More: NCERT Solutions For Class 11 Maths Binomial Theorem
Properties of Binomial Theorem
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Some of the Binomial Theorem Properties include:
- The sum of each term within the expanded (x+y)n is 1 + n.
- In each term, the sum of the indices of the two variables is n.
- The above expansion can be seen when the numbers x and y are complex numbers.
- It’s the coefficient that every term is equal (which means, equivalent in distance to one other) beginning till the point where they reach their conclusion.
- The binomial coefficients gradually increase to their highest value, following which they decrease.
Read Also: Difference between Power and Exponent
How to Apply Binomial Theorem?
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The following things need to be considered while applying the Binomial Theorem:
- By multiplying the initial term (a) with the exponents, the number n can further be reduced to zero.
- Now, the number of exponents for the term (b) is seen to range from zero to one hundred and fifty.
- Exponents are equivalent to the sum of their exponents. Similarly, it is the same for B and vice versa.
- For both the first and last terms, the function coefficients are one.
Applications of Binomial Theorem
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Some of the major applications of Binomial Theorem include:
- The binomial theorem is frequently used since the economy is reliant on both statistics and probability analysis.
- Binomial Theorem is often used in cases of higher mathematics and while calculating equations’ roots which contain higher power of magnitude.
- Binomial Theorem is also used in the Weather Forecast Services department.
- It is also widely used in Architecture, cost estimation, and other relevant fields.
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Things to Remember
- Binomial Theorem for Positive Integral Indices claims that “the total number of terms in the expansion is one more than the index”.
- Pascal’s Triangle defines that the triangular arrangement of numbers gives the coefficients in the expansion of a binomial expression.
- Binomial Theorem can be defined as a method of expanding an expression that is raised to a finite power.
- Pascal’s Triangle formula is = \({n \choose k}= {n-1 \choose k-1}+ {n-1 \choose k}\).
- The formula of Binomial theorem can be represented as: \((x+y)^n \sum_{k=0}^{n} {n \choose k} x^{n – k} y^k \).
Previous Year Questions
- The numerically greatest term in the binomial expansion of… (AP EAMCET - 2018)
- In the binomial expansion of (1+x)15 the coefficients of… (KCET - 2010)
- The total number of terms in the expansion of ( 1 + x )2n… (AMUEEE - 2012)
- Sum of last 30 coefficients in the binomial expansion of… (KEAM - 2018)
- Let tn denote the nth term in a binomial expansion. If… (KEAM)
- The coefficient of xn in the expansion of (1+ x)(1- x)n is… (AIEEE - 2004)
- The number of irrational terms in the expansion of… (WBJEE - 2019)
- The coefficient of x50 in the binomial expansion of… (JEE Main - 2014)
Sample Questions:
Ques: Using Binomial Theorem, evaluate (96)3. (3 marks)
Ans: 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.
It can be written that,
96 = 100−4
Therefore,
(96)3= (100−4)3
= 3C0(100)3 – 3C1(100)2(4)+ 3C2(100)(4)2- 3C3(4)3
=1000000−3(10000)(4)+3(100)(16)−64
=1000000−120000+4800−64
=884736
Therefore, (96)3= 884736
Ques: Expand the expression (x+1/x)6 (3 marks)
Ans: By using Binomial Theorem, the expression (x+1/x)6 can be expanded as

Ques: Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000 (3 marks)
Ans: By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 be obtained as
(1.1)10000 = (1+0.1)10000
= 10000C0 + 10000C1(1.1)+ other positive terms
= 1+10000 × 1.1+other positive terms
=1+11000+other positive terms > 1000
Hence, (1.1)10000 > 1000
Ques: Prove that n∑4=0(3)r nCr=4n (2 marks)
Ans: By Binomial Theorem,

Hence proved.
Ques: Find (a+b)4 - (a-b)4. Hence, evaluate (√3+√2)4- (√3-√2)4 (5 marks)
Ans: Using Binomial Theorem, the expressions (a+b)4 - (a-b)4 can be expanded as,

Ques: Write the general term in the expansion of: (x2-yx)12, x ≠ 0 (3 marks)
Ans: It is known that the general term Tr+1in the binomial expansion (a+b)n is given by

Ques: Find the 4th term in the expansion of (x-2y)12 . (3 marks)
Ans: We know, (r+1)n term, Tr+1

Ques: Find the middle terms in the expansions of (3 - x3/6)7 . (5 marks)
Ans: Thus,

Ques: Expand the expression 1-2x5 (3 marks)
Ans: By using the Binomial Theorem, the expression 1-2x5 can be expanded as,

Ques: Find the coefficient of x5 in (x+3)8 (3 marks)
Ans: We know, (r+1)th term, Tr+1 in the binomial expansion (a+b)n is given by
Tr+1=nCr an-rbr
Assuming that x2 occurs in the r+1th term of the expansion x+38,
We get,
Tr+1=aCr (x)8-r(3)r
Comparing the indices of x in x5 in Tr+1,
We obtain, r=3
Thus, the coefficient of x5 in 8C3(3)3 = 8! / 3!5! x 33= 8.7.6.5! / 3.2.5! x 33 = 1512
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