Binomial Theorem: JEE Advanced PYQ, 2007–2026

Answer: (A)

The coefficient of x^r is C(n, r)·2^(n−r)·(1/3)^r. Setting the x^7 and x^8 coefficients equal: C(n,7)·2^(n−7)/3^7 = C(n,8)·2^(n−8)/3^8. Dividing, C(n,8)/C(n,7) = 2·3 = 6. Since C(n,8)/C(n,7) = (n − 7)/8, we get n − 7 = 48, so n = 55.

Answer: (A)

In the second factor, the general term is C(9, r)(3/2)^(9−r)(−1/3)^r·x^(18−3r). The factor 1 needs x⁰, so 18 − 3r = 0 and r = 6. This gives C(9,6)(3/2)³(1/3)⁶ = 84·(27/8)·(1/729) = 7/18. The factor x needs x⁻¹, so 18 − 3r = −1, which has no integer solution. The factor 2x³ needs x⁻³, so r = 7. This gives 2·C(9,7)(3/2)²(−1/3)⁷ = 2·36·(9/4)·(−1/2187) = −2/27. The total is 7/18 − 2/27 = 21/54 − 4/54 = 17/54.

Answer: (D)

The coefficient of x^5 in (1+x)^k is C(k,5). Add over k = 5 to 10: 1 + 6 + 21 + 56 + 126 + 252 = 462. This agrees with the identity C(5,5)+C(6,5)+…+C(10,5) = C(11,6) = 462. The value 252 counts only (1+x)^10, and 461 omits the first term.

Answer: (A)

Use 8⁸³ = (1 + 7)⁸³ and 6⁸³ = (7 − 1)⁸³. Every term containing 7² or a higher power is divisible by 49, so only the first two terms of each expansion matter. 8⁸³ ≡ 1 + 83·7 = 582 ≡ 43 (mod 49). 6⁸³ = (7 − 1)⁸³ ≡ −1 + 83·7 = 580 ≡ 41 (mod 49), because the exponent 83 is odd and the linear term has sign (−1)⁸² = +1. The sum is 84 ≡ 35 (mod 49). Equivalently, the 1 and −1 cancel and the linear terms add to 2·83·7 = 1162 = 49·23 + 35.

Answer: (B)

Write 7^103 = 7·(7^2)^51 = 7·(50 − 1)^51. Expanding, (50 − 1)^51 = 50k + (−1)^51 = 50k − 1 for an integer k. So 7^103 = 350k − 7, and 350k is a multiple of 25. The remainder is −7 + 25 = 18.

Answer: (A)

Principle: (a + b)^n + (a − b)^n is an integer, because the odd powers of √5 cancel in the sum. Let g = (3 − √5)^7. Since 0 < 3 − √5 < 1, we have 0 < g < 1. Now I + f + g is an integer, so f + g is an integer in (0, 2). Since f is irrational and non-zero, f + g = 1, so 1 − f = g. Then (I + f)(1 − f) = (3+√5)^7 (3−√5)^7 = (9 − 5)^7 = 4^7 = 16384.

Answer: (B)

With f(x) = (1 + x + x²)⁵: f(1) = 3⁵ = 243 is the sum of all coefficients, and f(−1) = 1⁵ = 1 is even minus odd. The even-power coefficients sum to (243 + 1)/2 = 122.

Answer: (A)

The sum is a geometric series with ratio (1 + x) and 21 terms, so it equals ((1 + x)^21 − 1)/(1 + x − 1) = ((1 + x)^21 − 1)/x. The coefficient of x^8 in this quotient equals the coefficient of x^9 in (1 + x)^21 − 1. That is C(21, 9) = 293930. The same result follows from the hockey-stick identity: the sum over k of C(k, 8), for k from 8 to 20, equals C(21, 9).

Answer: (A)

The general term is T(r+1) = C(30,r)(3x)^(30−r) y^r in the first expansion, and C(30,r)(3x)^(30−r)(−y)^r in the second. In the difference, a term with y^r is multiplied by 1 − (−1)^r. For even r this is 0, and for odd r it is 2. So only r = 1, 3, 5, …, 29 survive, which is 15 terms.

Answer: (A)

Principle: substituting x = 1 gives the sum of all coefficients, and x = −1 gives the alternating sum. Adding them, P(1) + P(−1) = 2(a₀ + a₂ + … + a₂₀). Here P(1) = (1 − 1 + 1)^10 = 1 and P(−1) = (1 + 1 + 1)^10 = 3^10 = 59049. So the required sum is (1 + 59049)/2 = 29525.