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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 1-30 of 69 questions
1 Medium
1 of 69
Analyzing +4 | -1
123+12+225+12+22+327+…\frac{1^2}{3}+\frac{1^2+2^2}{5}+\frac{1^2+2^2+3^2}{7}+\ldots upto nn terms is equal to
View answer and explanation
Correct option: A
n(n+1)(n+2)18\frac{n(n+1)(n+2)}{18}
Option A is correct because the sum of the series up to n terms simplifies to n(n+1)(n+2)18\frac{{n(n+1)(n+2)}}{{18}}.
2 Medium
2 of 69
Analyzing +4 | -1
The value of sum in the nnth bracket of
View answer and explanation
Correct option: C
2n−1(2n+2n−1−1)2^{n-1}\left(2^n+2^{n-1}-1\right)
Option B is correct because the sum of the nth bracket can be found by analyzing the pattern and using the formula for the sum of an arithmetic series.
3 Medium
3 of 69
Analyzing +4 | -1
For a positive integer nn, Let a(n)=1+12+13+14+...+1(2n)−1.a\left( n \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{\left( {{2^n}} \right) - 1}}. then
View answer and explanation
Correct option: A
a(100)≤100a\left( {100} \right) \le 100
Option A (a(100)≤100a\left( {100} \right) \le 100) is correct because the sum of the series a(n)a(n) is bounded by nn, so a(100)a(100) is less than or equal to 100.
4 Medium
4 of 69
Application +4 | -1
Forces forming a couple are of magnitude 12 N12 {~N} each and the arm of the couple is 8 m8 {~m}. The force of an equivalent couple whose arm is 6 m6 {~m} is of magnitude
View answer and explanation
Correct option: B
16 N16 {~N}
Option B (16 N16 {~N}) is correct because the moment of a couple is constant, so equating the moments of the two couples gives the unknown force.
5 Medium
5 of 69
Application +4 | -1
By mathematical induction, 11⋅2⋅3+12⋅3⋅4+….+1n(n+1)(n+2)\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\ldots .+\frac{1}{n(n+1)(n+2)} is equal to
View answer and explanation
Correct option: B
n(n+3)4(n+1)(n+2)\frac{n(n+3)}{4(n+1)(n+2)}
Option B (n(n+3)4(n+1)(n+2)\frac{{n(n+3)}}{{4(n+1)(n+2)}}) is correct because the sum can be proven using mathematical induction.
6 Medium
6 of 69
Application +4 | -1
If 3x−42≥x+14−1\frac{3 x-4}{2} \geq \frac{x+1}{4}-1, then x∈x \in
View answer and explanation
Correct option: A
[1,∞)[1, \infty)
Option A ([1,∞)[1, \infty)) is correct because it represents the solution set where the inequality 3x−42≥x+14−1\frac{{3x-4}}{{2}} \geq \frac{{x+1}}{{4}}-1 is satisfied.
7 Medium
7 of 69
Analyzing +4 | -1
Sum of nnth bracket of
View answer and explanation
Correct option: B
(3n−1)32⋅4n−1\frac{\left(3^n-1\right)^3}{2 \cdot 4^{n-1}}
The correct answer is none of these because the provided options do not match the correct sum of the nth bracket of the given series.
8 Medium
8 of 69
Analyzing +4 | -1
1⋅22+2⋅32+3⋅42+…+n(n+1)212⋅2+22⋅3+32⋅4+…+n2(n+1)\frac{1 \cdot 2^2+2 \cdot 3^2+3 \cdot 4^2+\ldots+n(n+1)^2}{1^2 \cdot 2+2^2 \cdot 3+3^2 \cdot 4+\ldots+n^2(n+1)} is equal to
View answer and explanation
Correct option: A
3n+53n+1\frac{3 n+5}{3 n+1}
Option A (3n+53n+1\frac{{3n+5}}{{3n+1}}) is correct because the ratio of the summations simplifies to this expression after applying summation formulas and factoring.
9 Medium
9 of 69
Application +4 | -1
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\frac{3}{4}\right)^{6 x+10-x^2} < \frac{27}{64} is
View answer and explanation
Correct option: C
7
The correct answer is C (7) because solving the inequality leads to the range -1 < x < 7, which contains 7 integers.
10 Medium
10 of 69
Analyzing +4 | -1
If an=7+7+7+……a_n=\sqrt{7+\sqrt{7+\sqrt{7+\ldots \ldots}}} having n radical signs then by methods of mathematical induction which is true
View answer and explanation
Correct option: B
an<7 ∀ n≥1a_n < 7 \,\forall\, n \geq 1
Option B (an<7∀n≥1a_n < 7 \forall n \geq 1) is correct because by mathematical induction, it's shown that each term is less than 7.
11 Medium
11 of 69
Application +4 | -1
2⋅12+3⋅22+4⋅32+…+(n+1)n22 \cdot 1^2+3 \cdot 2^2+4 \cdot 3^2+\ldots+(n+1) n^2 is equal to
View answer and explanation
Correct option: D
n(n+1)(n+2)(3n+1)12\frac{n(n+1)(n+2)(3 n+1)}{12}
Option D (n(n+1)(n+2)(3n+1)12\frac{{n(n+1)(n+2)(3 n+1)}}{{12}}) is correct because it correctly sums the given series.
12 Medium
12 of 69
Application +4 | -1
The solution set of
2x−13≥(3x−24)−(2−x5)\frac{2 x-1}{3} \geq\left(\frac{3 x-2}{4}\right)-\left(\frac{2-x}{5}\right) is (−∞(-\infty, a]. The value of ' aa ' is
View answer and explanation
Correct option: A
2
Option A (2) is correct because solving the inequality results in x≤2x \leq 2, thus the value of 'a' is 2.
13 Medium
13 of 69
Application +4 | -1
A manufacturer has 600 litres of a 12%12 \% solution of acid. How many litres of a 30%30 \% acid solution must be added to it so that acid content in the resulting mixture will be more than 15%15 \% but less than 18%18 \% ?
View answer and explanation
Correct option: A
More than 120 litres but less than 300 litres
Option A (More than 120 litres but less than 300 litres) is correct because it represents the range of litres of 30% acid solution that must be added to achieve the desired acid concentration in the mixture.
14 Medium
14 of 69
Application +4 | -1
The set of real values of xx satisfying ∣x−1∣≤3|x-1| \leq 3 and ∣x−1∣⩾1\mid x-1| \geqslant 1 is
View answer and explanation
Correct option: C
[−2,0]∪[2,4][-2,0] \cup[2,4]
The correct answer is C ([-2,0] ∪\cup[2,4]) because it represents the intersection of the solution sets of the two inequalities.
15 Medium
15 of 69
Application +4 | -1
By the principle of induction ∀n∈N,32n\forall \mathrm{n} \in N, 3^{2 \mathrm{n}} when divided by 8 , leaves remainder
View answer and explanation
Correct option: D
1
Option D (1) is correct because 32n3^{{2n}} can be written as 9n9^n, and when divided by 8, the remainder is always 1.

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16 Medium
16 of 69
Analyzing +4 | -1
1.3⋅4+2⋅4⋅5+3⋅5⋅6+…3 \cdot 4+2 \cdot 4 \cdot 5+3 \cdot 5 \cdot 6+\ldots upto nn terms is equal to
View answer and explanation
Correct option: B
n(n+1)(3n2+23n+46)12\frac{n(n+1)\left(3 n^2+23 n+46\right)}{12}
Option B is correct because the sum of the series can be expressed as the sum of the product of consecutive numbers, which simplifies to the given expression.
17 Medium
17 of 69
Application +4 | -1
If nn is a positive integer, then 2⋅42n+1+33n+12 \cdot 4^{2 n+1}+3^{3 n+1} is divisible by :
View answer and explanation
Correct option: C
11
Option C (11) is correct because the expression 2⋅42n+1+33n+12 \cdot 4^{{2n+1}} + 3^{{3n+1}} is divisible by 11 for all positive integers n, which can be proven using mathematical induction.
18 Medium
18 of 69
Understanding +4 | -1
Let S(K)=1+3+5…+(2K−1)=3+K2S(K)=1+3+5 \ldots+(2 K-1)=3+K^2. Then which of the following is true
View answer and explanation
Correct option: B
S(K)⇒S(K+1)S(K) \Rightarrow S(K+1)
Option B (S(K)⇒S(K+1)S(K) \Rightarrow S(K+1)) is correct because even though S(1) is not true, the implication from S(K) to S(K+1) holds, meaning if S(K) were true, then S(K+1) would also be true.
19 Medium
19 of 69
Application +4 | -1
Statement p(n):8n≤2n−16p(n): 8 n \leq 2^n-16 is true for n>…….,n∈Nn>\ldots \ldots ., n \in \mathrm{N}.
View answer and explanation
Correct option: C
5
Option C (5) is correct because for n>5n > 5, the inequality 8n≤2n−168n \leq 2^n - 16 holds true, and it's the smallest integer value that satisfies this condition.
20 Medium
20 of 69
Application +4 | -1
Statement p(n):32n+1+2n−1,∀ n∈Np(n): 3^{2 n+1}+2^{n-1}, \forall \, n \in \mathrm{N} is divisible by
View answer and explanation
Correct option: C
7
Option C is correct. Using mathematical induction, it can be shown that the expression 3(2n+1) + 2(n-1) is divisible by 7 for all n in N.
21 Medium
21 of 69
Application +4 | -1
Solution of ∣2x−3∣<∣x+2∣|2 x-3| < |x+2| is
View answer and explanation
Correct option: B
(13,5)\left(\frac{1}{3}, 5\right)
Option B ((13,5)\left(\frac{1}{3}, 5\right)) is correct because solving the inequality ∣2x−3∣<∣x+2∣|2x-3| < |x+2| involves considering different cases based on the signs of 2x−32x-3 and x+2x+2, and the solution is the intersection of the solutions from each case.
22 Medium
22 of 69
Application +4 | -1
The least positive integer which makes n!<[n+12]nn ! < \left[\frac{n+1}{2}\right]^n is _____, where [ ] is the greatest integer function.
View answer and explanation
Correct option: C
3
Option C is correct. By testing values of n starting from 1, we find that n=3 is the least positive integer that satisfies the inequality n! < [(n+1)/2]^n.
23 Medium
23 of 69
Analyzing +4 | -1
Let S(k)=1+3+5....+(2k−1)=3+k2S\left( k \right) = 1 + 3 + 5.... + \left( {2k - 1} \right) = 3 + {k^2} Then, which of the following is true?
View answer and explanation
Correct option: B
S(k)⇒S(k+1)S\left( k \right) \Rightarrow S\left( {k + 1} \right)
Option B (S(k)⇒S(k+1)S(k) \Rightarrow S(k + 1)) is correct because if we assume S(k)S(k) is true, we can show that S(k+1)S(k+1) is also true.
24 Medium
24 of 69
Application +4 | -1
Let the number (22)2022+(2022)22(22)^{2022}+(2022)^{22} leave the remainder α\alpha when divided by 3 and β\beta when divided by 7 . Then (α2+β2)\left(\alpha^2+\beta^2\right) is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because by finding the remainders when divided by 3 and 7, and then calculating α2+β2\alpha^2 + \beta^2, the result is 5.
25 Medium
25 of 69
Analyzing +4 | -1
Let P(n):n2+n+1P\left( n \right):{n^2} + n + 1 is an even integer. If P(k)P\left( k \right) is assumed true ⇒P(k+1)\Rightarrow P\left( {k + 1} \right) is true. Therefore, P(n)P\left( n \right) is true
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the statement P(n) is not true for n=1, thus the principle of mathematical induction cannot be applied.
26 Medium
26 of 69
Application +4 | -1
For all n≥1n \geq 1, find 11.2+12.3+13.4+…..+1n(n+1)\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots . .+\frac{1}{n(n+1)}
View answer and explanation
Correct option: A
nn+1\frac{\mathrm{n}}{\mathrm{n}+1}
Option A (nn+1\frac{{\mathrm{{n}}}}{{\mathrm{{n}}+1}}) is correct because the sum can be simplified using partial fraction decomposition and mathematical induction.
27 Medium
27 of 69
Application +4 | -1
(23n−1)({2^{3n}} - 1) will be divisible by (∀n∈N)\left( {\forall n \in N} \right)
View answer and explanation
Correct option: C
7
Option C (7) is correct because (23n−1)({2^{3n}} - 1) is divisible by 7 for all n in N.
28 Medium
28 of 69
Analyzing +4 | -1
Let SK=1+2+…+KKS_{K}=\frac{1+2+\ldots+K}{K} and ∑j=1nSj2=nA(Bn2+Cn+D)\sum\limits_{{j=1}}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right), where A,B,C,D∈NA, B, C, D \in \mathbb{N} and AA has least value. Then
View answer and explanation
Correct option: D
A+BA+B is divisible by DD
Option D (A+BA+B is divisible by DD) is correct because when the given expression is simplified and the values of A, B, and D are substituted, the condition holds true.
29 Medium
29 of 69
Application +4 | -1
The number of integral roots of the equation ∣x−1∣+∣x+2∣−∣x−3∣=4|x-1|+|x+2|-|x-3|=4 is
View answer and explanation
Correct option: C
2
The correct answer is C (2) because solving the absolute value equation in different intervals yields two integral roots: -8 and 2.
30 Medium
30 of 69
Application +4 | -1
The sum of (12)+(12+22)+(12+22+32)\left(1^2\right)+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right) +…n+\ldots n brackets is equal to
View answer and explanation
Correct option: B
n(n+1)2(n+2)12\frac{n(n+1)^2(n+2)}{12}
Option B (n(n+1)2(n+2)12\frac{{n(n+1)^2(n+2)}}{{12}}) is correct because it correctly sums the series of sums of squares.
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