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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Mixed difficulty

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Showing 61-90 of 121 questions

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61 Easy
61 of 121
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.
62 Medium
62 of 121
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.
63 Easy
63 of 121
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.
64 Easy
64 of 121
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.
65 Easy
65 of 121
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.
66 Medium
66 of 121
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.
67 Medium
67 of 121
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.
68 Easy
68 of 121
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.
69 Easy
69 of 121
Application +4 | -1
The sum to nn terms of the series 13+33+53+…{1^3} + {3^3} + {5^3} + \ldots is
View answer and explanation
Correct option: B
n2(2n2−1){n^2}\left( {2{n^2} - 1} \right)
Option B (n2(2n2−1){n^2}(2{n^2} - 1)) is correct because it gives the sum of the series for n = 1, 2, etc.
70 Easy
70 of 121
Analyzing +4 | -1
Among the statements:

(S1) : 20232022−199920222023^{2022}-1999^{2022} is divisible by 8.

(S2) : 13(13)n−11n−1313(13)^{n}-11 n-13 is divisible by 144 for infinitely many n∈Nn \in \mathbb{N}
View answer and explanation
Correct option: B
Only (S1) is correct
Option B (Only (S1) is correct) is correct because the first statement is divisible by 8, while the second statement is not divisible by 144 for infinitely many n.
71 Medium
71 of 121
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.
72 Easy
72 of 121
Application +4 | -1
For all n∈N,4n−3n−1n \in N,{4^n} - 3n - 1 is divisible by
View answer and explanation
Correct option: C
9
Option C (9) is correct because 4n−3n−1{4^n} - 3n - 1 is divisible by 9 for all n∈Nn \in N. This can be verified by testing a few values of n or using mathematical induction.
73 Medium
73 of 121
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.
74 Medium
74 of 121
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.
75 Easy
75 of 121
Application +4 | -1
Using mathematical induction, then numbers an ′s{a_n}{\,^\prime }s are defined by
a0=1,an+1=3n2+n+an,(n≥0){a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \ge 0} \right) Then, an{a_n} is equal to
View answer and explanation
Correct option: B
n3−n2+1{n^3} - {n^2} + 1
Option B (n3−n2+1{n^3} - {n^2} + 1) is correct because it satisfies the given recurrence relation and initial condition.

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76 Medium
76 of 121
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, ...
77 Medium
77 of 121
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.
78 Medium
78 of 121
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.
79 Medium
79 of 121
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.
80 Easy
80 of 121
Application +4 | -1
The nthnth terms of the series 3+7+13+21+…3 + 7 + 13 + 21 + \ldots is
View answer and explanation
Correct option: C
n2+n+1{n^2} + n + 1
Option C (n2+n+1{n^2} + n + 1) is correct because it generates the correct terms of the series when n = 1, 2, 3, and so on.
81 Medium
81 of 121
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.
82 Medium
82 of 121
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.
83 Medium
83 of 121
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}.
84 Medium
84 of 121
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.
85 Easy
85 of 121
Application +4 | -1
(1)+(1+2)+(1+2+3)+…n(1)+(1+2)+(1+2+3)+\ldots n brackets is equal to
View answer and explanation
Correct option: A
n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}
Option A (n(n+1)(n+2)6\frac{{n(n+1)(n+2)}}{{6}}) is correct because it correctly sums the series of sums of natural numbers.
86 Medium
86 of 121
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}.
87 Easy
87 of 121
Application +4 | -1
Using mathematical induction, then numbers an′s{a_n}^\prime s are defined by a0=1,an+1=3n2+n+an,(n≥0){a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \ge 0} \right) then, an{a_n} is equal to
View answer and explanation
Correct option: B
n3−n2+1{n^3} - {n^2} + 1
Option A (n3+n2+1{n^3} + {n^2} + 1) is incorrect. Option B (n3−n2+1{n^3} - {n^2} + 1) is correct because it satisfies the given recurrence relation and initial condition.
88 Medium
88 of 121
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.
89 Medium
89 of 121
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.
90 Medium
90 of 121
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.
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