The sum to $n$ terms of the series ${1^3} + {3^3} + {5^3} + \ldots $ is

${n^2}\left( {{n^2} - 1} \right)$

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

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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Easy

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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 31-51 of 51 questions

31 Easy

31 of 51

Application +4 | -1

The sum to

n

terms of the series

{1^3} + {3^3} + {5^3} + \ldots

View answer and explanation

Correct option: B

{n^2}\left( {2{n^2} - 1} \right)

Option B (

{n^2}(2{n^2} - 1)

) is correct because it gives the sum of the series for n = 1, 2, etc.

32 Easy

32 of 51

Analyzing +4 | -1

Among the statements:

(S1) :

2023^{2022}-1999^{2022}

is divisible by 8.

(S2) :

13(13)^{n}-11 n-13

is divisible by 144 for infinitely many

n \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.

33 Easy

33 of 51

Application +4 | -1

For all

n \in N,{4^n} - 3n - 1

is divisible by

View answer and explanation

Correct option: C

Option C (9) is correct because

{4^n} - 3n - 1

is divisible by 9 for all

n \in N

. This can be verified by testing a few values of n or using mathematical induction.

34 Easy

34 of 51

Application +4 | -1

Using mathematical induction, then numbers

{a_n}{\,^\prime }s

are defined by

{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \ge 0} \right)

Then,

{a_n}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it satisfies the given recurrence relation and initial condition.

35 Easy

35 of 51

Application +4 | -1

The

nth

terms of the series

3 + 7 + 13 + 21 + \ldots

View answer and explanation

Correct option: C

{n^2} + n + 1

Option C (

{n^2} + n + 1

) is correct because it generates the correct terms of the series when n = 1, 2, 3, and so on.

36 Easy

36 of 51

Application +4 | -1

(1)+(1+2)+(1+2+3)+\ldots n

brackets is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)(n+2)}{6}

Option A (

\frac{{n(n+1)(n+2)}}{{6}}

) is correct because it correctly sums the series of sums of natural numbers.

37 Easy

37 of 51

Application +4 | -1

Using mathematical induction, then numbers

{a_n}^\prime s

are defined by

{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \ge 0} \right)

then,

{a_n}

is equal to

View answer and explanation

Correct option: B

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

Option A (

{n^3} + {n^2} + 1

) is incorrect. Option B (

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

) is correct because it satisfies the given recurrence relation and initial condition.

38 Easy

38 of 51

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.

39 Easy

39 of 51

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.

40 Easy

40 of 51

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.

41 Easy

41 of 51

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

42 Easy

42 of 51

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.

43 Easy

43 of 51

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.

44 Easy

44 of 51

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.

45 Easy

45 of 51

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.

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46 Easy

46 of 51

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.

47 Easy

47 of 51

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.

48 Easy

48 of 51

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.

49 Easy

49 of 51

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.

50 Easy

50 of 51

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

51 Easy

51 of 51

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.

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