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Binomial Theorem & its Simple Applications: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Binomial Theorem (January 2026) Q1: Given below are two statements : Statement I : 25 13 + 20 13 + 8 13 + 3 13 is divisible by 7. Statement II : The integral part of (7 + 4v3) 25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : (a) Statement I is false but Statement II is true (b) Both Statement I and Statement II are false (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Ans: (c) Sol: Statement I: 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Key concept : If n is odd, xn + yn = (x + y)(xn ?¹ - xn ?²y + xn ?³y² - · · · - xyn ?² + yn ?¹) So, xn + yn is divisible by x + y. In the given statement : 25¹³ + 3¹³ is divisible by 25 + 3 = 28. 25¹³ + 3¹³ is also divisible by 7 because 28 is divisible by 7. So, 25¹³ + 3¹³ = 7 · I 1..........(1) ( I 1 is a positive integer). Similarly : 20¹³ + 8¹³ is divisible by 20 + 8 = 28. 20¹³ + 8¹³ is divisible by 7. So, 20¹³ + 8¹³ = 7 · I 2 ..........(2) ( I 2 is a positive integer). Add (1) & (2) : 25¹³ + 3¹³ + 20¹³ + 8¹³ = 7(I 1 + I 2) So, 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Statement I is true. Statement II: The integral part of (7 + 4v3)² 5 is an odd number. Let X = (7 + 4v3)² 5. Let I be the integral part and f be the fractional part ( $0 Adding the two expansions: (7 + 4v3)² 5 + (7 - 4v3)² 5 = I + f + f' Using the binomial expansion (a + b)n + (a - b)n = 2[an + (n 2)an ?²b² + . . .], we see that the sum is always an even integer because of the factor of 2. I + f + f' = Even Integer Since I is an integer, f + f' must also be an integer. Given 0 < f < 1 and 0 < f' < 1, the only possible integer value for their sum is f + f' = 1. I + 1 = Even Integer I = Even Integer - 1 = Odd Integer Page 2 JEE Main Previous Year Questions (2021-2026): Binomial Theorem (January 2026) Q1: Given below are two statements : Statement I : 25 13 + 20 13 + 8 13 + 3 13 is divisible by 7. Statement II : The integral part of (7 + 4v3) 25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : (a) Statement I is false but Statement II is true (b) Both Statement I and Statement II are false (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Ans: (c) Sol: Statement I: 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Key concept : If n is odd, xn + yn = (x + y)(xn ?¹ - xn ?²y + xn ?³y² - · · · - xyn ?² + yn ?¹) So, xn + yn is divisible by x + y. In the given statement : 25¹³ + 3¹³ is divisible by 25 + 3 = 28. 25¹³ + 3¹³ is also divisible by 7 because 28 is divisible by 7. So, 25¹³ + 3¹³ = 7 · I 1..........(1) ( I 1 is a positive integer). Similarly : 20¹³ + 8¹³ is divisible by 20 + 8 = 28. 20¹³ + 8¹³ is divisible by 7. So, 20¹³ + 8¹³ = 7 · I 2 ..........(2) ( I 2 is a positive integer). Add (1) & (2) : 25¹³ + 3¹³ + 20¹³ + 8¹³ = 7(I 1 + I 2) So, 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Statement I is true. Statement II: The integral part of (7 + 4v3)² 5 is an odd number. Let X = (7 + 4v3)² 5. Let I be the integral part and f be the fractional part ( $0 Adding the two expansions: (7 + 4v3)² 5 + (7 - 4v3)² 5 = I + f + f' Using the binomial expansion (a + b)n + (a - b)n = 2[an + (n 2)an ?²b² + . . .], we see that the sum is always an even integer because of the factor of 2. I + f + f' = Even Integer Since I is an integer, f + f' must also be an integer. Given 0 < f < 1 and 0 < f' < 1, the only possible integer value for their sum is f + f' = 1. I + 1 = Even Integer I = Even Integer - 1 = Odd Integer The integral part I is indeed an odd number. Statement II is True. Q2: The sum of the coefficients of x 499 and x 500 in (1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 998 + … + x 1000 is: (a) 1002 C 501 (b) 1001 C 501 (c) 1000 C 501 (d) 1002 C 500 Ans: (d) Sol: Let S = (1 + x)¹ ° ° ° + x(1 + x) ? ? ? + x²(1 + x) ? ? 8 + · · · + x¹ ° ° ° Notice that this is a Geometric Progression (GP) with: First term (a): (1 + x)¹ ° ° ° Common ratio (r) : x/(1+x) Number of terms (n): 1001 The sum of a GP is given by Substituting our values: Simplify the denominator: S = (1 + x)¹ ° °¹ - x¹ ° °¹ Now we need the coefficients of x 4 ? ? and x5 ° ° in the simplified expression (1 + x)¹ ° °¹ - x¹ ° °¹. Since x¹ ° °¹ only affects the term with power 1001, we only look at (1 + x)¹ ° °¹. In the binomial expansion (1 + x)n Coefficient of x? is nC? Coefficient of x 4 ? ? in S : ¹ ° °¹C 4 9 9 Coefficient of x 5 ° ° in S : ¹ ° °¹C 5 0 0 (The term x¹ ° °¹ does not contribute.) Sum of coefficients ¹ ° °¹C 4 9 9 + ¹ ° °¹C 5 0 0 = ¹ ° °²C 5 0 0 [Using pascal's identity nC? + nC? ? 1 = n ?¹C? ? 1] Q3: Let up to 13 terms. If 13 S = 2 k /n! , k ? N , then n + k is equal to Page 3 JEE Main Previous Year Questions (2021-2026): Binomial Theorem (January 2026) Q1: Given below are two statements : Statement I : 25 13 + 20 13 + 8 13 + 3 13 is divisible by 7. Statement II : The integral part of (7 + 4v3) 25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : (a) Statement I is false but Statement II is true (b) Both Statement I and Statement II are false (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Ans: (c) Sol: Statement I: 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Key concept : If n is odd, xn + yn = (x + y)(xn ?¹ - xn ?²y + xn ?³y² - · · · - xyn ?² + yn ?¹) So, xn + yn is divisible by x + y. In the given statement : 25¹³ + 3¹³ is divisible by 25 + 3 = 28. 25¹³ + 3¹³ is also divisible by 7 because 28 is divisible by 7. So, 25¹³ + 3¹³ = 7 · I 1..........(1) ( I 1 is a positive integer). Similarly : 20¹³ + 8¹³ is divisible by 20 + 8 = 28. 20¹³ + 8¹³ is divisible by 7. So, 20¹³ + 8¹³ = 7 · I 2 ..........(2) ( I 2 is a positive integer). Add (1) & (2) : 25¹³ + 3¹³ + 20¹³ + 8¹³ = 7(I 1 + I 2) So, 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Statement I is true. Statement II: The integral part of (7 + 4v3)² 5 is an odd number. Let X = (7 + 4v3)² 5. Let I be the integral part and f be the fractional part ( $0 Adding the two expansions: (7 + 4v3)² 5 + (7 - 4v3)² 5 = I + f + f' Using the binomial expansion (a + b)n + (a - b)n = 2[an + (n 2)an ?²b² + . . .], we see that the sum is always an even integer because of the factor of 2. I + f + f' = Even Integer Since I is an integer, f + f' must also be an integer. Given 0 < f < 1 and 0 < f' < 1, the only possible integer value for their sum is f + f' = 1. I + 1 = Even Integer I = Even Integer - 1 = Odd Integer The integral part I is indeed an odd number. Statement II is True. Q2: The sum of the coefficients of x 499 and x 500 in (1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 998 + … + x 1000 is: (a) 1002 C 501 (b) 1001 C 501 (c) 1000 C 501 (d) 1002 C 500 Ans: (d) Sol: Let S = (1 + x)¹ ° ° ° + x(1 + x) ? ? ? + x²(1 + x) ? ? 8 + · · · + x¹ ° ° ° Notice that this is a Geometric Progression (GP) with: First term (a): (1 + x)¹ ° ° ° Common ratio (r) : x/(1+x) Number of terms (n): 1001 The sum of a GP is given by Substituting our values: Simplify the denominator: S = (1 + x)¹ ° °¹ - x¹ ° °¹ Now we need the coefficients of x 4 ? ? and x5 ° ° in the simplified expression (1 + x)¹ ° °¹ - x¹ ° °¹. Since x¹ ° °¹ only affects the term with power 1001, we only look at (1 + x)¹ ° °¹. In the binomial expansion (1 + x)n Coefficient of x? is nC? Coefficient of x 4 ? ? in S : ¹ ° °¹C 4 9 9 Coefficient of x 5 ° ° in S : ¹ ° °¹C 5 0 0 (The term x¹ ° °¹ does not contribute.) Sum of coefficients ¹ ° °¹C 4 9 9 + ¹ ° °¹C 5 0 0 = ¹ ° °²C 5 0 0 [Using pascal's identity nC? + nC? ? 1 = n ?¹C? ? 1] Q3: Let up to 13 terms. If 13 S = 2 k /n! , k ? N , then n + k is equal to (a) 50 (b) 52 (c) 49 (d) 51 Ans: (c) Sol: multiply in numerator and denominator by 26 ! Lets find 13th term lower subscript are in A.P with, common difference = 2 It is given that substitute value of S Page 4 JEE Main Previous Year Questions (2021-2026): Binomial Theorem (January 2026) Q1: Given below are two statements : Statement I : 25 13 + 20 13 + 8 13 + 3 13 is divisible by 7. Statement II : The integral part of (7 + 4v3) 25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : (a) Statement I is false but Statement II is true (b) Both Statement I and Statement II are false (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Ans: (c) Sol: Statement I: 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Key concept : If n is odd, xn + yn = (x + y)(xn ?¹ - xn ?²y + xn ?³y² - · · · - xyn ?² + yn ?¹) So, xn + yn is divisible by x + y. In the given statement : 25¹³ + 3¹³ is divisible by 25 + 3 = 28. 25¹³ + 3¹³ is also divisible by 7 because 28 is divisible by 7. So, 25¹³ + 3¹³ = 7 · I 1..........(1) ( I 1 is a positive integer). Similarly : 20¹³ + 8¹³ is divisible by 20 + 8 = 28. 20¹³ + 8¹³ is divisible by 7. So, 20¹³ + 8¹³ = 7 · I 2 ..........(2) ( I 2 is a positive integer). Add (1) & (2) : 25¹³ + 3¹³ + 20¹³ + 8¹³ = 7(I 1 + I 2) So, 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Statement I is true. Statement II: The integral part of (7 + 4v3)² 5 is an odd number. Let X = (7 + 4v3)² 5. Let I be the integral part and f be the fractional part ( $0 Adding the two expansions: (7 + 4v3)² 5 + (7 - 4v3)² 5 = I + f + f' Using the binomial expansion (a + b)n + (a - b)n = 2[an + (n 2)an ?²b² + . . .], we see that the sum is always an even integer because of the factor of 2. I + f + f' = Even Integer Since I is an integer, f + f' must also be an integer. Given 0 < f < 1 and 0 < f' < 1, the only possible integer value for their sum is f + f' = 1. I + 1 = Even Integer I = Even Integer - 1 = Odd Integer The integral part I is indeed an odd number. Statement II is True. Q2: The sum of the coefficients of x 499 and x 500 in (1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 998 + … + x 1000 is: (a) 1002 C 501 (b) 1001 C 501 (c) 1000 C 501 (d) 1002 C 500 Ans: (d) Sol: Let S = (1 + x)¹ ° ° ° + x(1 + x) ? ? ? + x²(1 + x) ? ? 8 + · · · + x¹ ° ° ° Notice that this is a Geometric Progression (GP) with: First term (a): (1 + x)¹ ° ° ° Common ratio (r) : x/(1+x) Number of terms (n): 1001 The sum of a GP is given by Substituting our values: Simplify the denominator: S = (1 + x)¹ ° °¹ - x¹ ° °¹ Now we need the coefficients of x 4 ? ? and x5 ° ° in the simplified expression (1 + x)¹ ° °¹ - x¹ ° °¹. Since x¹ ° °¹ only affects the term with power 1001, we only look at (1 + x)¹ ° °¹. In the binomial expansion (1 + x)n Coefficient of x? is nC? Coefficient of x 4 ? ? in S : ¹ ° °¹C 4 9 9 Coefficient of x 5 ° ° in S : ¹ ° °¹C 5 0 0 (The term x¹ ° °¹ does not contribute.) Sum of coefficients ¹ ° °¹C 4 9 9 + ¹ ° °¹C 5 0 0 = ¹ ° °²C 5 0 0 [Using pascal's identity nC? + nC? ? 1 = n ?¹C? ? 1] Q3: Let up to 13 terms. If 13 S = 2 k /n! , k ? N , then n + k is equal to (a) 50 (b) 52 (c) 49 (d) 51 Ans: (c) Sol: multiply in numerator and denominator by 26 ! Lets find 13th term lower subscript are in A.P with, common difference = 2 It is given that substitute value of S n + K = 25 + 24 = 49 Q4: The sum of all possible values of n ? N, so that the coefficients of x, x 2 and x 3 in the expansion of (1 + x 2 ) 2 (1 + x) n , are in arithmetic progression is: (a) 12 (b) 9 (c) 3 (d) 7 Ans: (b) Sol: (1 + x²)²(1 + xn) (1 + x 4 + 2x²)(1 + x)n Coeff of x = nC 1 Coeff of x² = nC 2 + 2 · nC 0 Coeff of x³ = nC 3 + 2 · nC 1 ? Coeff of x, x², x³ ? AP ? 2(nC 2 + 2) = nC 1 + nC 3 + 2n 2nC 2 + 4 = nC 3 + 3n ? n³ - 9n² + 26n - 24 = 0 ? n = 2, 3, 4 ? Sum of all value of n is 9 Q5: The value of (a) (b) (c) Page 5 JEE Main Previous Year Questions (2021-2026): Binomial Theorem (January 2026) Q1: Given below are two statements : Statement I : 25 13 + 20 13 + 8 13 + 3 13 is divisible by 7. Statement II : The integral part of (7 + 4v3) 25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : (a) Statement I is false but Statement II is true (b) Both Statement I and Statement II are false (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Ans: (c) Sol: Statement I: 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Key concept : If n is odd, xn + yn = (x + y)(xn ?¹ - xn ?²y + xn ?³y² - · · · - xyn ?² + yn ?¹) So, xn + yn is divisible by x + y. In the given statement : 25¹³ + 3¹³ is divisible by 25 + 3 = 28. 25¹³ + 3¹³ is also divisible by 7 because 28 is divisible by 7. So, 25¹³ + 3¹³ = 7 · I 1..........(1) ( I 1 is a positive integer). Similarly : 20¹³ + 8¹³ is divisible by 20 + 8 = 28. 20¹³ + 8¹³ is divisible by 7. So, 20¹³ + 8¹³ = 7 · I 2 ..........(2) ( I 2 is a positive integer). Add (1) & (2) : 25¹³ + 3¹³ + 20¹³ + 8¹³ = 7(I 1 + I 2) So, 25¹³ + 20¹³ + 8¹³ + 3¹³ is divisible by 7. Statement I is true. Statement II: The integral part of (7 + 4v3)² 5 is an odd number. Let X = (7 + 4v3)² 5. Let I be the integral part and f be the fractional part ( $0 Adding the two expansions: (7 + 4v3)² 5 + (7 - 4v3)² 5 = I + f + f' Using the binomial expansion (a + b)n + (a - b)n = 2[an + (n 2)an ?²b² + . . .], we see that the sum is always an even integer because of the factor of 2. I + f + f' = Even Integer Since I is an integer, f + f' must also be an integer. Given 0 < f < 1 and 0 < f' < 1, the only possible integer value for their sum is f + f' = 1. I + 1 = Even Integer I = Even Integer - 1 = Odd Integer The integral part I is indeed an odd number. Statement II is True. Q2: The sum of the coefficients of x 499 and x 500 in (1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 998 + … + x 1000 is: (a) 1002 C 501 (b) 1001 C 501 (c) 1000 C 501 (d) 1002 C 500 Ans: (d) Sol: Let S = (1 + x)¹ ° ° ° + x(1 + x) ? ? ? + x²(1 + x) ? ? 8 + · · · + x¹ ° ° ° Notice that this is a Geometric Progression (GP) with: First term (a): (1 + x)¹ ° ° ° Common ratio (r) : x/(1+x) Number of terms (n): 1001 The sum of a GP is given by Substituting our values: Simplify the denominator: S = (1 + x)¹ ° °¹ - x¹ ° °¹ Now we need the coefficients of x 4 ? ? and x5 ° ° in the simplified expression (1 + x)¹ ° °¹ - x¹ ° °¹. Since x¹ ° °¹ only affects the term with power 1001, we only look at (1 + x)¹ ° °¹. In the binomial expansion (1 + x)n Coefficient of x? is nC? Coefficient of x 4 ? ? in S : ¹ ° °¹C 4 9 9 Coefficient of x 5 ° ° in S : ¹ ° °¹C 5 0 0 (The term x¹ ° °¹ does not contribute.) Sum of coefficients ¹ ° °¹C 4 9 9 + ¹ ° °¹C 5 0 0 = ¹ ° °²C 5 0 0 [Using pascal's identity nC? + nC? ? 1 = n ?¹C? ? 1] Q3: Let up to 13 terms. If 13 S = 2 k /n! , k ? N , then n + k is equal to (a) 50 (b) 52 (c) 49 (d) 51 Ans: (c) Sol: multiply in numerator and denominator by 26 ! Lets find 13th term lower subscript are in A.P with, common difference = 2 It is given that substitute value of S n + K = 25 + 24 = 49 Q4: The sum of all possible values of n ? N, so that the coefficients of x, x 2 and x 3 in the expansion of (1 + x 2 ) 2 (1 + x) n , are in arithmetic progression is: (a) 12 (b) 9 (c) 3 (d) 7 Ans: (b) Sol: (1 + x²)²(1 + xn) (1 + x 4 + 2x²)(1 + x)n Coeff of x = nC 1 Coeff of x² = nC 2 + 2 · nC 0 Coeff of x³ = nC 3 + 2 · nC 1 ? Coeff of x, x², x³ ? AP ? 2(nC 2 + 2) = nC 1 + nC 3 + 2n 2nC 2 + 4 = nC 3 + 3n ? n³ - 9n² + 26n - 24 = 0 ? n = 2, 3, 4 ? Sum of all value of n is 9 Q5: The value of (a) (b) (c) (d) Ans: (b) Sol: Q6: Let C r denote the coefficient of x r in the binomial expansion of (1 + x) n , n ?N, 0 = r = n. If , then the value of equals. (a) 675 (b) 580 (c) 525 (d) 650 Ans: (a) Sol:Read More
FAQs on Binomial Theorem & its Simple Applications: JEE Main Previous Year Questions (2021-2026)
| 1. What is the Binomial Theorem? |  |
Ans. The Binomial Theorem is a fundamental theorem in algebra that provides a formula for the expansion of powers of a binomial expression. It states that (a + b)ⁿ can be expanded as the sum of terms in the form C(n, k) a^(n-k) b^k, where C(n, k) is the binomial coefficient, representing the number of ways to choose k elements from n elements. The theorem is expressed as (a + b)ⁿ = ∑ (C(n, k) a^(n-k) b^k) for k = 0 to n.
| 2. What are binomial coefficients and how are they calculated? | |
Ans. Binomial coefficients, denoted as C(n, k) or nCk, represent the number of ways to choose k elements from a set of n elements without regard to the order of selection. They can be calculated using the formula C(n, k) = n! / (k!(n-k)!), where n! (n factorial) is the product of all positive integers up to n. These coefficients appear in the expansion of a binomial expression according to the Binomial Theorem.
| 3. How can the Binomial Theorem be applied to find specific coefficients in an expansion? | |
Ans. To find specific coefficients in the expansion of (a + b)ⁿ, one can use the Binomial Theorem. For example, to find the coefficient of a^k in the expansion of (a + b)ⁿ, identify the term corresponding to k by using the formula C(n, k) a^(n-k) b^k. The coefficient of a^k is then given by C(n, k) multiplied by the appropriate power of b. This method allows for the determination of any particular term in the expansion.
| 4. Can the Binomial Theorem be used for negative integers or fractions? | |
Ans. Yes, the Binomial Theorem can be extended to negative integers and fractions through the use of the generalized binomial series. For a negative integer n, the expansion can be expressed as (1 + x)⁻ⁿ = ∑ (C(-n, k) x^k) for k = 0 to ∞, where C(-n, k) is defined using the formula C(-n, k) = (-1)ᵏ C(n + k - 1, k). For fractional values, the series converges under certain conditions, typically when |x| < 1.
| 5. What are some simple applications of the Binomial Theorem in problem-solving? | |
Ans. Simple applications of the Binomial Theorem include calculating probabilities in binomial distributions, determining the coefficients of terms in polynomial expansions, and solving combinatorial problems. It can also be used to simplify expressions in algebra, such as finding the value of (x + y)⁴ or to derive useful identities such as the binomial identities for summing powers and calculating expected values in statistics.
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