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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Easy

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Practice format and chapter quiz

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 51 questions
1 Easy
1 of 51
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.
2 Easy
2 of 51
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).
3 Easy
3 of 51
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.
4 Easy
4 of 51
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.
5 Easy
5 of 51
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}.
6 Easy
6 of 51
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.
7 Easy
7 of 51
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.
8 Easy
8 of 51
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.
9 Easy
9 of 51
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.
10 Easy
10 of 51
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.
11 Easy
11 of 51
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.
12 Easy
12 of 51
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.
13 Easy
13 of 51
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.
14 Easy
14 of 51
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.
15 Easy
15 of 51
Understanding +4 | -1
A student was asked to prove a statement P(n)P(n) by induction. He proved that P(k+1)P(k+1) is true whenever P(k)P(k) is true for all k>5∈Nk>5 \in \mathrm{N} and also that P(5)P(5) is true. On the basis of this he could conclude that P(n)P(n) is true
View answer and explanation
Correct option: C
for all n≥5n \geq 5
Option C (for all n≥5n \geq 5) is correct because the base case is P(5) and the inductive step holds for k > 5, meaning it holds for all n greater than or equal to 5.

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16 Easy
16 of 51
Analyzing +4 | -1
Let P(n)P\left( n \right) denotes the statement that n2+n{n^2} + n is odd. It is seen that P(n)⇒P(n+1),  P(n)P\left( n \right) \Rightarrow P\left( {n + 1} \right),\;P\left( n \right) is true for all
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because n2+n{n^2} + n is always even, not odd, so the statement P(n)P(n) is never true.
17 Easy
17 of 51
Application +4 | -1
10n+3(4n+2)+5{10^n} + 3\left( {{4^{n + 2}}} \right) + 5 is divisible by (n∈N)\left( {n \in N} \right)
View answer and explanation
Correct option: C
9
Option C (9) is correct because when n=2n = 2 is substituted into the expression 10n+3(4n+2)+5{10^n} + 3\left( {{4^{n + 2}}} \right) + 5, the result is 873, which is divisible by 9.
18 Easy
18 of 51
Application +4 | -1
If x2n−1+y2n−1{x^{2n - 1}} + {y^{2n - 1}} is divisible by x+yx + y, if nn is
View answer and explanation
Correct option: A
A positive integer
Option A (A positive integer) is correct because x2n−1+y2n−1{x^{2n - 1}} + {y^{2n - 1}} is divisible by x+yx + y for all positive integers n.
19 Easy
19 of 51
Analyzing +4 | -1
Statement I:- P(n) is a true statement ∀n⩾1;&n∉N\forall n \geqslant 1;\& n \notin N if P(n)P(n) is true for n=t&t+1n = t\& t + 1 where t∈Nt \in N equal to set of natural numbers
Statement II:- Any statement P(n)P(n) is true ∀n∈N\forall n \in N by the principle of finite mathematical
induction. Where N=N = set of natural numbers
View answer and explanation
Correct option: C
I is true, II is true and is a correct explanation for I
I is false, II is true. The principle of mathematical induction applies only to natural numbers, not to all real numbers greater than or equal to 1.
20 Easy
20 of 51
Application +4 | -1
If nn is a positive integer, then 52n+2−24n−25{5^{2n + 2}} - 24n - 25 isdivisible by
View answer and explanation
Correct option: D
576
Option D (576) is correct because when n=1, the expression evaluates to 576, and when n=2, it evaluates to a multiple of 576.
21 Easy
21 of 51
Understanding +4 | -1
The number an−bn(a,b{a^n} - {b^n}(a,b are distinct rational numbers and n∈Nn \in N) is always divisible by
View answer and explanation
Correct option: A
a−ba - b
Option A (a−ba - b) is correct because an−bn{a^n} - {b^n} can be factored as (a−b)(an−1+an−2b+...+bn−1)(a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1}), making it always divisible by a−ba-b.
22 Easy
22 of 51
Application +4 | -1
p(n):22np(n): 2^{2^n} has _____ at unit's place for n>1n>1.
View answer and explanation
Correct option: B
6
Option B is correct. For n > 1, 2(2^n) always has 6 at the unit's place, as demonstrated by induction.
23 Easy
23 of 51
Application +4 | -1
For all n∈N,3n5+5n3+7nn \in N,3{n^5} + 5{n^3} + 7n is divisible by
View answer and explanation
Correct option: D
15
Option D (15) is correct because for all n∈N,3n5+5n3+7nn \in N,3{n^5} + 5{n^3} + 7n is divisible by 15. This can be verified by testing a few values of n.
24 Easy
24 of 51
Application +4 | -1
If nn is a positive integer, then 52n+2−24n−25{5^{2n + 2}} - 24n - 25 isdivisible by
View answer and explanation
Correct option: D
576
Option D (576) is correct because the expression 52n+2−24n−25{5^{2n + 2}} - 24n - 25 is always divisible by 576 for positive integers n.
25 Easy
25 of 51
Application +4 | -1
For all n∈N,72n−48n−1n \in N,{7^{2n}} - 48n - 1 is divisible by
View answer and explanation
Correct option: D
2304
Option D (2304) is correct because 72n−48n−1{7^{2n}} - 48n - 1 is divisible by 482{48^2} which equals 2304.
26 Easy
26 of 51
Application +4 | -1
For all n∈N,33n−26n−1n \in N,{3^{3n}} - {26^n} - 1 is divisible by
View answer and explanation
Correct option: D
676
Option D (676) is correct because 33n−26n−1{3^{3n}} - {26^n} - 1 can be expressed as a series divisible by 262{26^2}, which is 676.
27 Easy
27 of 51
Application +4 | -1
10n+3(4n+2)+5{10^n} + 3\left( {{4^{n + 2}}} \right) + 5 is divisible by (n∈N)\left( {n \in N} \right)
View answer and explanation
Correct option: C
9
Option C (9) is correct because the expression is divisible by 9 for all n belonging to N, as can be shown by induction or by testing for n=1.
28 Easy
28 of 51
Application +4 | -1
For all n∈N,52n−1n \in N,{5^{2n}} - 1 is divisible by
View answer and explanation
Correct option: C
24
Option C (24) is correct because 52n−1{5^{2n}} - 1 can be expressed as (1+24)n−1(1 + 24)^n - 1, which is always divisible by 24 for all n∈Nn \in N.
29 Easy
29 of 51
Analyzing +4 | -1
For n∈N,  10n−2≥81nn \in N,\;{10^{n - 2}} \ge 81n is
View answer and explanation
Correct option: B
n≥5n \ge 5
Option B (n≥5n \ge 5) is correct because the inequality holds true for n=5n=5 and can be proven by mathematical induction for all n≥5n \ge 5.
30 Easy
30 of 51
Analyzing +4 | -1
(1)+(2+3)+(4+5+6)+…n(1)+(2+3)+(4+5+6)+\ldots n brackets is equal to
View answer and explanation
Correct option: A
n(n+1)(n2+n+2)8\frac{n(n+1)\left(n^2+n+2\right)}{8}
Option A is correct because the sum of the series up to n brackets can be calculated using the formula for the sum of the first n natural numbers and the sum of the squares of the first n natural numbers.
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