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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Medium

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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 61-69 of 69 questions

61 Medium

61 of 69

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.

62 Medium

62 of 69

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.

63 Medium

63 of 69

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}

64 Medium

64 of 69

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.

65 Medium

65 of 69

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.

66 Medium

66 of 69

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.

67 Medium

67 of 69

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

68 Medium

68 of 69

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.

69 Medium

69 of 69

Application +4 | -1

49^n+16 n+\lambda

is divisible by 64 for all

n \in \mathrm{N}

, then the least negative value of

\lambda

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because when n=1,

49^n + 16n + \lambda

becomes

49 + 16 + \lambda = 65 + \lambda

, which is divisible by 64 if

\lambda = -1

Revision notes and key takeaways

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