Binomial Theorem (Algebra, Teaching Exam)Total Questions: 7561. Sum of all the coefficients of expansion (x+1)ⁿ is: [PGT 2002](a) n(n+1)2ⁿ⁻²(b) n(n+1)2ⁿ⁺²(c) 0(d) 2ⁿCorrect Answer: (d) 2ⁿSolution:62. The number of permutations of n distinct things taken r together, in which 3 particular things must occur, is: [PGT 2002](а) ⁿP(b) ⁿ⁻³Pᵣ₋₃(c) ⁿ⁻³Pₙ r!(d) None of theseCorrect Answer: (b) ⁿ⁻³Pᵣ₋₃Solution:The number of permutations of n distinct things taken r together, in which 3 particular things must occur is = ⁿ⁻³Pᵣ₋₃63. Number of terms in the expansion of (a+b+c)²⁵ is [Rajasthan TGT 2016, 2013, UP PGT 2019,16](a) 351(b) 325(c) 625(d) 338Correct Answer: (a) 351Solution:64. In the expansion of (1+px)ⁿ,n,p ∈ N the coefficients of x and x² are 8 and 24 respectively, then [Rajasthan TGT 2015](a) n =3,p = 2(b) n =4,p = 2(c) n = 4,p = 3(d) n = 5,p = 3Correct Answer: (b) n =4,p = 2Solution: 65. The middle term in the expansion of (x+a)²ⁿ; is [Rajasthan TGT 2015](a)(b)(c)(d)Correct Answer: (d)Solution:66. The binomial expansion of (1+x)⁻¹ is valid only when [(GIC) 2015](a) -1≤x ≤1(b) x≤1(c) -1<x <1(d) x≥1Correct Answer: (c) -1<x <1Solution:Binomial expansion of (1+x)⁻¹ is valid only when -1<x<167. Solve the following equation? [Rajasthan TGT 2013] (a)(b)(c)(d)Correct Answer: (c)Solution:68. In the expansion of (x+a)ⁿ, P is the sum of odd terms and Q is even the sum of the terms. Then what is the value of (x+a)²ⁿ-(x-a)²ⁿ. [TGT 2013](a) P+Q(b) P-Q(c) PQ(d) 4PQCorrect Answer: (d) 4PQSolution: 69. Solve the following equtaion? [TGT 2011](a)(b)(c)(d)Correct Answer: (a)Solution:70. Solve the following equation? [TGT 2010](a) 252(b) -252(c) 253(d) -253Correct Answer: (b) -252Solution:Submit Quiz« Previous12345678Next »