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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Mixed difficulty

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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 121 questions
1 Medium
1 of 121
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 Easy
2 of 121
Analyzing +4 | -1
If P(n):2+4+6…+(2n),  n∈N,P\left( n \right):2 + 4 + 6 \ldots + \left( {2n} \right),\;n \in N, then
P(k)=k(k+1)+2impliesP(k)=(k+1)(k+2)+2\begin{array}{l} P\left( k \right) = k\left( {k + 1} \right) + 2\mathrm{implies}\\ P\left( k \right) = \left( {k + 1} \right)\left( {k + 2} \right) + 2 \end{array}
is true for all k∈N.k \in N. So, statement P(n)=n(n+1)+2P\left( n \right) = n\left( {n + 1} \right) + 2 is true for
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the base case P(1)P(1) is not true, meaning mathematical induction cannot be applied.
3 Medium
3 of 121
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.
4 Medium
4 of 121
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.
5 Easy
5 of 121
Application +4 | -1
The greatest positive integer, which divides n(n+1)(n+2)(n+3)n(n+1)(n+2)(n+3) for all n∈Nn \in N, is
View answer and explanation
Correct option: C
24
Option C (24) is correct because the product of four consecutive integers is always divisible by 4! (which is 24).
6 Medium
6 of 121
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.
7 Easy
7 of 121
Analyzing +4 | -1
(1)+(1+3)+(1+3+5)+…n(1)+(1+3)+(1+3+5)+\ldots n brackets is equal to
View answer and explanation
Correct option: D
n(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}
Option D is correct because the sum of the terms in the nth bracket is the sum of the first n odd numbers, which is n squared, and the sum of the first n squares is given by the formula n(n+1)(2n+1)/6.
8 Easy
8 of 121
Application +4 | -1
Consider the statement : P(n):n2−n+41  is  prime.P(n):{n^2} - n + 41\mathrm{ }\,\,\mathrm{is}\,\,\mathrm{ prime}\mathrm{.} Then, which one of the following is true?
View answer and explanation
Correct option: A
Both P(3) and P(5) are true.
Option C is correct because after substitution and simplification, the integral evaluates to the given expression.
9 Easy
9 of 121
Application +4 | -1
If a,ba, b and cc are the sides of a △ABC\triangle A B C, then a+b−c\sqrt{a}+\sqrt{b}-\sqrt{c} is always
View answer and explanation
Correct option: B
positive
Option B (positive) is correct because the triangle inequality states that the sum of any two sides of a triangle is greater than the third side, which implies a+b>c\sqrt{a} + \sqrt{b} > \sqrt{c}.
10 Easy
10 of 121
Application +4 | -1
For all n∈N,  49n+16n−1n \in N,\;{49^n} + 16n - 1 is divisible by
View answer and explanation
Correct option: A
64
Option A (64) is correct because 49n+16n−1{49^n} + 16n - 1 is divisible by 64 for all n∈Nn \in N. This can be verified by testing a few values of n.
11 Medium
11 of 121
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.
12 Medium
12 of 121
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.
13 Medium
13 of 121
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.
14 Medium
14 of 121
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.
15 Medium
15 of 121
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.

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16 Medium
16 of 121
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.
17 Medium
17 of 121
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.
18 Medium
18 of 121
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.
19 Easy
19 of 121
Application +4 | -1
For all n∈N,2.42n+1+33n+1n \in N,{2.4^{2n + 1}} + {3^{3n + 1}} is divisible by
View answer and explanation
Correct option: D
11
Option D (11) is correct because when n=1, the expression 2.42n+1+33n+1{2.4^{2n + 1}} + {3^{3n + 1}} equals 209, which is divisible by 11.
20 Easy
20 of 121
Analyzing +4 | -1
If P(n):2+4+6+…+(2n),  n∈N,P\left( n \right):2 + 4 + 6 + \ldots + \left( {2n} \right),\;n \in N, then
P(k)=k(k+1)+2 impliesP(k+1)=(k+1)(k+2)+2\begin{array}{l} P\left( k \right) = k\left( {k + 1} \right) + 2\,\mathrm{implies}\\ P\left( {k + 1} \right) = \left( {k + 1} \right)\left( {k + 2} \right) + 2 \end{array}
is true for all k∈N.k \in N. So, statement P(n)=n(n+1)+2P\left( n \right) = n\left( {n + 1} \right) + 2 is true for
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the statement P(n) = n(n+1) + 2 is not true for any n >= 1, as P(1) is not equal to 1(1+1) + 2 = 4.
21 Medium
21 of 121
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.
22 Easy
22 of 121
Application +4 | -1
For each n∈N,32n−1n \in N,{3^{2n}} - 1 is divisible by
View answer and explanation
Correct option: A
8
Option A (8) is correct because 32n−1{3^{2n}} - 1 is divisible by 8 for all n∈Nn \in N. This can be proven using mathematical induction.
23 Easy
23 of 121
Application +4 | -1
For each, n∈N,102n−1+1n \in N,{10^{2n - 1}} + 1 is divisible by
View answer and explanation
Correct option: A
11
Option A (11) is correct because 102n−1+1{10^{2n - 1}} + 1 is divisible by 11 for all n in N.
24 Medium
24 of 121
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.
25 Easy
25 of 121
Understanding +4 | -1
For every natural number n,n(n+1)n, n(n+1) is always
View answer and explanation
Correct option: A
Even
Option A (Even) is correct because the product of two consecutive integers is always divisible by 2, hence even.
26 Medium
26 of 121
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.
27 Easy
27 of 121
Application +4 | -1
For all positive integral values of n,  32n−2n+1n,\;{3^{2n}} - 2n + 1 is divisible by
View answer and explanation
Correct option: A
2
Option A is correct because substituting n = 1 into the expression yields a number divisible by 2.
28 Medium
28 of 121
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.
29 Easy
29 of 121
Application +4 | -1
For all n∈N,n3+2nn \in N,{n^3} + 2n is divisible by
View answer and explanation
Correct option: A
3
Option A (3) is correct because for n=1, 2, 3, the expression n3 + 2n is divisible by 3.
30 Easy
30 of 121
Application +4 | -1
3+13+29+51+79+…3 + 13 + 29 + 51 + 79 + \ldots to nn terms =
View answer and explanation
Correct option: C
n3+2n2{n^3} + 2{n^2}
Option C is correct because the formula n³ + 2n² generates the correct sum for n=1, 2, and 3.
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  • ✓ This set contains 121 single-correct-answer MCQs on MATHEMATICAL INDUCTION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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