If $x = {\log _{{2^2}}}2 + {\log _{{2^3}}}{2^2} + {\log _{{2^4}}}{2^3} + \; \ldots .\; + {\log _{{2^{n + 1}}}}{2^n},$ then

$x \ge {\left( {\frac{1}{{n + 1}}} \right)^{1/n}}$

$x \ge {\left( {\frac{n}{{n + 1}}} \right)^{1/n}}$

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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 91-120 of 121 questions

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91 Easy

91 of 121

Application +4 | -1

The smallest positive integer

n

for which

n! < {\left( {\;\frac{{n + 1}}{2}} \right)^n}

holds, is

View answer and explanation

Correct option: B

Option B (2) is correct because when n=2, the inequality

n! < {\left( {\;\frac{{n + 1}}{2}} \right)^n}

holds true, and it's the smallest positive integer for which it's valid.

92 Medium

92 of 121

Application +4 | -1

5^{40}

is divided by 11 , then remainder is

\alpha

and when

2^{2003}

is divided by 17 , then remainder is

\beta

, then the value of

\beta-\alpha

View answer and explanation

Correct option: A

Option A (7) is correct because the remainders when

5^{40}

and

2^{2003}

are divided by 11 and 17 respectively are 1 and 8, and their difference is 7.

93 Easy

93 of 121

Application +4 | -1

n

is a positive integer, then

{n^3} + 2n

is divisible by

View answer and explanation

Correct option: D

Option D (3) is correct because

{n^3} + 2n

is divisible by 3 for all positive integers n. This can be shown by testing a few values or using modular arithmetic.

94 Easy

94 of 121

Understanding +4 | -1

Let

\mathrm{P}(n):{ }^" 2^n < (1 \times 2 \times 3 \times \ldots \times n) "

. Then the smallest positive integer for which

\mathrm{P}(n)

is true is

View answer and explanation

Correct option: A

Option A (4) is correct because it's the smallest positive integer for which

2^n < (1 \times 2 \times 3 \times ... \times n)

holds true.

95 Medium

95 of 121

Application +4 | -1

The area and perimeter of a rectangle are

A

P

respectively. Then

P

and

A

satisfy the inequality

View answer and explanation

Correct option: D

P^2 \geq 16 A

Option D (

P^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.

96 Easy

96 of 121

Understanding +4 | -1

\left( n \right):1 + 3 + 5 + ... + \left( {2n - 1} \right) = {n^2}

View answer and explanation

Correct option: A

True for all

n \in N

Option A (True for all

n \in N

) is correct because the sum of the first n odd natural numbers is always equal to

n^2

97 Medium

97 of 121

Application +4 | -1

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

Option A is correct because substituting n=1 into the expression yields the correct sum, and the formula can be proven by induction.

98 Medium

98 of 121

Application +4 | -1

{2^{3n}} - 7n - 1

is divisible by

View answer and explanation

Correct option: C

Option C (49) is correct because

{2^{3n}} - 7n - 1

is divisible by 49, which can be proven using mathematical induction or by rewriting the expression using the binomial theorem.

99 Medium

99 of 121

Application +4 | -1

The set of real values of

x

satisfying ||

x-1|-1| \mid \leq 1

View answer and explanation

Correct option: A

[-1,3]

The correct answer is A ([-1,3]) because solving the nested absolute value inequality leads to this interval.

100 Medium

100 of 121

Application +4 | -1

x

satisfies the inequalities

x+7 < 2 x+3

and

2 x+4 < 5 x+3

, then

x

lies in the interval

View answer and explanation

Correct option: C

(4, \infty)

Option C (

(4, \infty)

) is correct because it represents the interval where both inequalities

x+7 < 2x+3

and

2x+4 < 5x+3

are satisfied.

101 Medium

101 of 121

Application +4 | -1

For all

\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

\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.

102 Easy

102 of 121

Application +4 | -1

\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=1

, then

\mathrm{ab}+\mathrm{bc}+\mathrm{ca}

lies in the interval :

View answer and explanation

Correct option: A

\left[-\frac{1}{2}, 1\right]

Option A is correct because the interval for ab+bc+ca is derived from the given condition a²+b²+c²=1 and the properties of squares.

103 Easy

103 of 121

Application +4 | -1

For every natural number

n, 3^{2 n+2}-8 n-9

is divisible by___.

View answer and explanation

Correct option: A

Option A (16) is correct because substituting n=2 into the expression results in a number divisible by 16.

104 Hard

104 of 121

Analyzing +4 | -1

x = {\log _{{2^2}}}2 + {\log _{{2^3}}}{2^2} + {\log _{{2^4}}}{2^3} + \; \ldots .\; + {\log _{{2^{n + 1}}}}{2^n},

then

View answer and explanation

Correct option: B

x \ge n{\left( {\frac{1}{{n + 1}}} \right)^{1/n}}

Option B,

x \ge n{\left( {\frac{1}{{n + 1}}} \right)^{1/n}}

, is correct because it is derived by applying AM >= GM inequality to the given logarithmic expression.

105 Easy

105 of 121

Application +4 | -1

1^3+2^3+3^3+\ldots+50^3=m^2

, then

m=

View answer and explanation

Correct option: A

1275

Option A is correct. The sum of cubes from 1 to 50 is equal to (50*51/2)², which equals m², therefore m = 1275.

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106 Easy

106 of 121

Application +4 | -1

n \in N,

then

{3^{2n}} + 7

is divisible by

View answer and explanation

Correct option: B

Option B (8) is correct because

{3^{2n}} + 7

is divisible by 8 for n = 1, 2, 3, etc.

107 Easy

107 of 121

Understanding +4 | -1

The product of three consecutive natural numbers is divisible by

View answer and explanation

Correct option: C

Option C (6) is correct because at least one of any three consecutive natural numbers must be divisible by 2 and one must be divisible by 3, hence their product is divisible by 6.

108 Easy

108 of 121

Application +4 | -1

The remainder when

{5^{99}}

is divided by 13, is

View answer and explanation

Correct option: B

Option B (8) is correct because when

{5^{99}}

is divided by 13, the remainder is 8, which can be found by using modular arithmetic.

109 Medium

109 of 121

Analyzing +4 | -1

1+4+10+19+\ldots+\frac{\left(3 n^2-3 n+2\right)}{2}

is equal to

View answer and explanation

Correct option: D

\frac{n\left(n^2+1\right)}{2}

Option D (

\frac{{n(n^2+1)}}{{2}}

) is correct because the sum of the series is derived by breaking it into simpler summations and applying standard summation formulas.

110 Medium

110 of 121

Understanding +4 | -1

P\left( n \right)

is a statement

\left( {n \in N} \right)

such that, if

P\left( k \right)

is true,

P\left( {k + 1} \right)

is true for

k \in N

, then

p\left( n \right)

is true

View answer and explanation

Correct option: D

Nothing can be said

Option D (Nothing can be said) is correct because the base case P(1) must be proven true for the statement to be true for all n.

111 Medium

111 of 121

Understanding +4 | -1

\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{n+1}\right)

is equal to

View answer and explanation

Correct option: B

\frac{n}{n+1}

Option A (

\frac{1}{n+1}

) is incorrect. Option B (

\frac{n}{n+1}

) is correct because the given expression simplifies to

\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot ... \cdot \frac{n}{n+1}

, where most terms cancel out, leaving

\frac{1}{n+1}

112 Easy

112 of 121

Analyzing +4 | -1

x\left( {{x^{n - 1}} - n{\alpha ^{n - 1}}} \right) + {\alpha ^n}\left( {n - 1} \right)

is divisible by

{\left( {x - \alpha } \right)^2}

for

View answer and explanation

Correct option: C

All

n \in N

Option C (All

n \in N

) is correct because the expression is divisible by

(x-\alpha)^2

for all natural numbers n, as proven by induction.

113 Medium

113 of 121

Application +4 | -1

1 \cdot 2^2+2 \cdot 3^2+3 \cdot 4^2+\ldots+n(n+1)^2

is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)(n+2)(3 n+5)}{12}

Option A (

\frac{{n(n+1)(n+2)(3 n+5)}}{{12}}

) is correct because it correctly sums the given series.

114 Medium

114 of 121

Understanding +4 | -1

For every positive integer

n, 7^n-3^n

is divisible by

View answer and explanation

Correct option: C

Option C (4) is correct because

7^n - 3^n

can be shown to be divisible by 4 for all positive integers n using mathematical induction.

115 Medium

115 of 121

Application +4 | -1

1 \cdot 3+3 \cdot 5+5 \cdot 7+\ldots

upto

n

terms is equal to

View answer and explanation

Correct option: C

\frac{n\left(4 n^2+6 n-1\right)}{3}

Option C 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 it.

116 Easy

116 of 121

Application +4 | -1

n

is a positive integer, then

{n^3} + 2n

is divisible by

View answer and explanation

Correct option: D

Option D (3) is correct because when substituting positive integers into the expression

{n^3} + 2n

, the result is always divisible by 3.

117 Medium

117 of 121

Application +4 | -1

p(n): 49^n+24^{n-1}

is divisible by _____ for

n \in \mathrm{N}

View answer and explanation

Correct option: D

Option D (25) is correct because

49^n + 24^{n-1}

is always divisible by 25 for all natural numbers

n

118 Easy

118 of 121

Application +4 | -1

{a_1} = 1

and

{a_n} = n{a_{n - 1}}

for all positive integer

n \ge 2,

then

{a_5}

is equal to

View answer and explanation

Correct option: B

120

Option B (120) is correct because by iteratively applying the recursive formula

{a_n} = n{a_{n - 1}}

with

{a_1} = 1

, we find

{a_5} = 120

119 Medium

119 of 121

Understanding +4 | -1

The value of the series

\frac{1}{2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots .+\frac{1}{n(n+1)}

is equal to

View answer and explanation

Correct option: B

\frac{n}{n+1}

Option B (

\frac{n}{n+1}

) is correct because the series telescopes, leaving only the first and last terms after cancellation.

120 Easy

120 of 121

Application +4 | -1

m,n

are any two odd positive integer with

n < m

, then the largest positive integers which divides all the numbers of the type

{m^2} - {n^2}

View answer and explanation

Correct option: C

Option C (8) is correct because the expression m² - n² can be simplified to 8k, where k is an integer, indicating divisibility by 8.

Revision notes and key takeaways

Review MATHEMATICAL INDUCTION in your textbook or class notes, then use this set to check your understanding.

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