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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 69 questions
31 Medium
31 of 69
Application +4 | -1
A company manufactures cassettes. Its cost and revenue functions are C(x)=26000+30x\mathrm{C}(\mathrm{x})=26000+30 \mathrm{x} and R(x)\mathrm{R}(\mathrm{x}) =43x=43 x, respectively, where xx is the number of cassettes produced and sold in a week. The number of cassettes must be sold by the company to realise some profit, is
View answer and explanation
Correct option: A
More than 2000
Option A (More than 2000) is correct because the profit function, calculated as Revenue - Cost, must be greater than zero. Solving the inequality leads to x > 2000.
32 Medium
32 of 69
Analyzing +4 | -1
141⋅3+243⋅5+345⋅7+…+n4(2n+1)(2n−1)\frac{1^4}{1 \cdot 3}+\frac{2^4}{3 \cdot 5}+\frac{3^4}{5 \cdot 7}+\ldots+\frac{n^4}{(2 n+1)(2 n-1)} is equal to
View answer and explanation
Correct option: B
n(n+1)(n2+n+1)6(2n+1)\frac{n(n+1)\left(n^2+n+1\right)}{6(2 n+1)}
Option B (n(n+1)(n2+n+1)6(2n+1)\frac{{n(n+1)(n^2+n+1)}}{{6(2n+1)}}) is correct because it represents the sum of the given series after applying algebraic manipulations and summation formulas.
33 Medium
33 of 69
Analyzing +4 | -1
34+1516+6364+…n\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\ldots n terms is equal to
View answer and explanation
Correct option: A
n−134n−13n-\frac{1}{3} 4^n-\frac{1}{3}
Option B (n+134−n−13n+\frac{{1}}{{3}} 4^{-n}-\frac{{1}}{{3}}) is correct because substituting n=1 into the expression yields the first term of the series, 3/4.
34 Medium
34 of 69
Application +4 | -1
If n∈Nn \in N, then 11n+2+122n+1{11^{n + 2}} + {12^{2n + 1}} is divisible by
View answer and explanation
Correct option: C
133
Option C (133) is correct because when n=1, 11n+2+122n+1{11^{n + 2}} + {12^{2n + 1}} equals 3059, which is divisible by 133.
35 Medium
35 of 69
Application +4 | -1
1+4+13+40+…1+4+13+40+\ldots upto nn terms is equal to
View answer and explanation
Correct option: B
(3n+1−2n−3)4\frac{\left(3^{n+1}-2 n-3\right)}{4}
Option B is correct because substituting n=1 into the expression yields the correct sum, and the formula is derived by finding the general term and summing the geometric progression.
36 Medium
36 of 69
Analyzing +4 | -1
If (1)+(2+3+4)+(5+6+7+8+9)+…(1)+(2+3+4)+(5+6+7+8+9)+\ldots, then the sum of terms in the nnth bracket is
View answer and explanation
Correct option: A
(n−1)3+n3(n-1)^3+n^3
Option A is correct because the sum of terms in the nth bracket can be expressed as the sum of cubes of (n-1) and n.
37 Medium
37 of 69
Application +4 | -1
For all n∈N,3.52n+1+23n+1\mathrm{n} \in \mathrm{N}, 3.5^{2 \mathrm{n}+1}+2^{3 \mathrm{n}+1} is divisible by
View answer and explanation
Correct option: B
17
Option B (17) is the correct answer because when n=1 and n=2, the expression is divisible by 17.
38 Medium
38 of 69
Application +4 | -1
For all n∈N,41n−14n\mathrm{n} \in \mathrm{N}, 41^{\mathrm{n}}-14^{\mathrm{n}} is a multiple of___.
View answer and explanation
Correct option: A
27
Option A (27) is correct because 41n−14n41^n - 14^n is always a multiple of 27 for all natural numbers n, as can be proven by induction.
39 Medium
39 of 69
Analyzing +4 | -1
For natural number n,  2n(n−1)!<nnn,\;{2^n}\left( {n - 1} \right)! < {n^n}, if
View answer and explanation
Correct option: B
n>2n > 2
Option B (n>2n > 2) is correct because the inequality 2n(n−1)!<nn{2^n}(n - 1)! < n^n holds true for n>2n > 2.
40 Medium
40 of 69
Understanding +4 | -1
If P(n)P\left( n \right) is a statement such that P(3)P\left( 3 \right) is true. Assuming P(k)P\left( k \right) is true ⇒P(k+1)\Rightarrow P\left( {k + 1} \right) is true for all k≥3,k \ge 3, then P(n)P\left( n \right) is true
View answer and explanation
Correct option: B
For n≥3n \ge 3
Option B (For n≥3n \ge 3) is correct because the base case P(3) is true, and the inductive step shows that if P(k) is true, then P(k+1) is true for all k >= 3.
41 Medium
41 of 69
Application +4 | -1
The solution set of the inequality 5x+2>(125)1/x5^{x+2}>\left(\frac{1}{25}\right)^{1 / x}
View answer and explanation
Correct option: D
(0,∞)(0, \infty)
Option D ((0,∞)(0, \infty)) is correct because it represents the solution set where the inequality 5x+2>(125)1/x5^{x+2} > (\frac{{1}}{{25}})^{1/x} is satisfied.
42 Medium
42 of 69
Application +4 | -1
The nthn\mathrm{th} term of the series 4+14+30+52+80+114+…4 + 14 + 30 + 52 + 80 + 114 + \ldots is
View answer and explanation
Correct option: C
3n2+n3{n^2} + n
Option C (3n2+n3{n^2} + n) is correct because it generates the terms of the series when n = 1, 2, 3, ...
43 Medium
43 of 69
Analyzing +4 | -1
The inequality n!>2n−1n! > {2^{n - 1}} is true for
View answer and explanation
Correct option: A
n>2n > 2
Option A (n>2n > 2) is correct because the inequality n!>2n−1n! > 2^{n-1} holds true for n=3n=3 and can be proven by induction for all n>2n > 2.
44 Medium
44 of 69
Application +4 | -1
For any n∈N\mathrm{n} \in \mathbf{N}, the value of the expression 2+2+.....+2n−roots\mathop {\sqrt {2 + \sqrt {2 +.....+ \sqrt 2 } } }\limits_{\mathrm{n - roots}} is
View answer and explanation
Correct option: A
2cos⁡(π2n+1)2 \cos \left(\frac{\pi}{2^{n+1}}\right)
Option A (2cos⁡(π2n+1)2 \cos \left(\frac{{\pi}}{{2^{{n+1}}}}\right)) is correct because the expression can be simplified using trigonometric identities and mathematical induction.
45 Medium
45 of 69
Application +4 | -1
The solution set of (x)2+(x+1)2=25(x)^2+(x+1)^2=25, where (x)(x) is the least integer greater than or exual to xx, is
View answer and explanation
Correct option: B
(−5,−4]∪(2,3](-5,-4] \cup(2,3]
Option B ((-5,-4] ∪\cup(2,3]) is correct because it represents the solution set for x that satisfies the given equation involving the least integer function.
46 Medium
46 of 69
Application +4 | -1
Solution of ∣x−1∣≥∣x−3∣|x-1| \geq|x-3| is
View answer and explanation
Correct option: B
x≥2x \geq 2
The correct answer is B (x≥2x \geq 2) because the solution represents all points where the distance from x to 1 is greater than or equal to the distance from x to 3.
47 Medium
47 of 69
Application +4 | -1
Fractional part of the number 4202215\frac{4^{2022}}{15} is equal to
View answer and explanation
Correct option: D
115\frac{1}{15}
Option D (115\frac{1}{15}) is correct because 420224^{2022} is congruent to 1 modulo 15, making the fractional part 1/15.
48 Medium
48 of 69
Analyzing +4 | -1
If x1=3{x_1} = 3 and xn+1=2+xn,  n≥1{x_{n + 1}} = \sqrt {2 + {x_n}} ,\;n \ge 1, then lim⁡n→∞xn\lim\limits_{n \to \infty } {x_n} is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because the limit of the recursively defined sequence can be found by solving x=2+xx = \sqrt{2+x}.
49 Medium
49 of 69
Application +4 | -1
1⋅2+2⋅3+3⋅4+…1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots upto nn terms is equal to
View answer and explanation
Correct option: B
n(n+1)(n+2)3\frac{n(n+1)(n+2)}{3}
Option B (n(n+1)(n+2)3\frac{{n(n+1)(n+2)}}{{3}}) is correct because it correctly sums the given series.
50 Medium
50 of 69
Understanding +4 | -1
11⋅2+12⋅3+13⋅4+…n\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots n terms is equal to
View answer and explanation
Correct option: B
nn+1\frac{n}{n+1}
Option B (nn+1\frac{n}{n+1}) is correct because the given series is a telescoping series that simplifies to nn+1\frac{n}{n+1}.
51 Medium
51 of 69
Application +4 | -1
For all n∈N\mathrm{n} \in \mathrm{N}, 1+11+2+11+2+3+….+11+2+3+……+n1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots .+\frac{1}{1+2+3+\ldots \ldots+n} is equal to
View answer and explanation
Correct option: D
2nn+1\frac{2 n}{n+1}
Option D is correct because the sum can be simplified using the formula for the sum of natural numbers and then proven by induction.
52 Medium
52 of 69
Application +4 | -1
If ∑r=1kr=12(n2+11n+30)\sum\limits_{{r=1}}^k r=\frac{1}{2}\left(n^2+11 n+30\right), then k=k=
View answer and explanation
Correct option: D
n+5n+5
Option D (n+5n+5) is correct because by substituting k=n+5k=n+5 into the summation formula, we obtain the given expression.
53 Medium
53 of 69
Analyzing +4 | -1
If P(n):3n<n!,  n∈NP\left( n \right):{3^n} < n!,\;n \in N, then P(n)P\left( n \right) is true
View answer and explanation
Correct option: B
For n≥7n \ge 7
Option B (For n≥7n \ge 7) is correct because the inequality 3n<n!3^n < n! holds true for n=7n=7 and can be proven by induction for all n≥7n \ge 7.
54 Medium
54 of 69
Application +4 | -1
If 5405^{40} is divided by 11 , then remainder is α\alpha and when 220032^{2003} is divided by 17 , then remainder is β\beta, then the value of β−α\beta-\alpha is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the remainders when 5405^{40} and 220032^{2003} are divided by 11 and 17 respectively are 1 and 8, and their difference is 7.
55 Medium
55 of 69
Application +4 | -1
The area and perimeter of a rectangle are AA PP respectively. Then PP and AA satisfy the inequality
View answer and explanation
Correct option: D
P2≥16AP^2 \geq 16 A
Option D (P2≥16AP^2 \geq 16A) is correct because it represents the relationship between the perimeter and area of a rectangle, derived from the inequality between arithmetic and geometric means.
56 Medium
56 of 69
Application +4 | -1
1⋅1!+2⋅2!+3⋅3!+…+n⋅n!1 \cdot 1 !+2 \cdot 2 !+3 \cdot 3 !+\ldots+n \cdot n ! is equal to
View answer and explanation
Correct option: A
(n+1)!−1(n+1) !-1
Option A is correct because substituting n=1 into the expression yields the correct sum, and the formula can be proven by induction.
57 Medium
57 of 69
Application +4 | -1
23n−7n−1{2^{3n}} - 7n - 1 is divisible by
View answer and explanation
Correct option: C
49
Option C (49) is correct because 23n−7n−1{2^{3n}} - 7n - 1 is divisible by 49, which can be proven using mathematical induction or by rewriting the expression using the binomial theorem.
58 Medium
58 of 69
Application +4 | -1
The set of real values of xx satisfying ||x−1∣−1∣∣≤1x-1|-1| \mid \leq 1 is
View answer and explanation
Correct option: A
[−1,3][-1,3]
The correct answer is A ([-1,3]) because solving the nested absolute value inequality leads to this interval.
59 Medium
59 of 69
Application +4 | -1
If xx satisfies the inequalities x+7<2x+3x+7 < 2 x+3 and 2x+4<5x+32 x+4 < 5 x+3, then xx lies in the interval
View answer and explanation
Correct option: C
(4,∞)(4, \infty)
Option C ((4,∞)(4, \infty)) is correct because it represents the interval where both inequalities x+7<2x+3x+7 < 2x+3 and 2x+4<5x+32x+4 < 5x+3 are satisfied.
60 Medium
60 of 69
Application +4 | -1
For all n∈N,1.3+2.32+3.33+….+n.3n\mathrm{n} \in \mathrm{N}, 1.3+2.3^2+3.3^3+\ldots .+\mathrm{n} .3^{\mathrm{n}} is equal to
View answer and explanation
Correct option: B
(2n−1)3n+1+34\frac{(2 n-1) 3^{n+1}+3}{4}
Option B is correct. The sum of the series can be found using mathematical induction and algebraic manipulation.
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