JEE Mathematics Principle of Mathematical Induction MCQs Set 06

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Chapter-wise Objective Questions: Mathematical Induction

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Multiple Choice Questions

Question 1: The inequality n! > 2n−1 is true for
(a) n > 2
(b) n ∈ N
(c) n > 3
(d) None of these
Show Answer & Explanation

Answer: (a) n > 2

Explanation:
1. n = 1: 1 > 1 is false; n = 2: 2 > 2 is false; n = 3: 6 > 4 is true.
2. Assuming k! > 2k−1 with k ≥ 3, (k+1)! = (k+1)k! > (k+1)2k−1 > 2k, so the inequality holds for all n > 2.

Teacher's Note:
1. Test small values of n and then use induction from the first true case.

Question 2: The statement P(n): 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! − 1 is
(a) True for all n > 1
(b) Not true for any n
(c) True for all n ∈ N
(d) None of these
Show Answer & Explanation

Answer: (c) True for all n ∈ N

Explanation:
1. n = 1: 1 = 2! − 1 holds.
2. If true for n = k, adding (k+1)(k+1)! gives (k+1)! − 1 + (k+1)(k+1)! = (k+2)! − 1, so it holds for k + 1.
3. Hence it is true for all n ∈ N.

Teacher's Note:
1. The step uses (k+1)! (1 + k + 1) = (k+2)!.

Question 3: If n ∈ N, then 11n+2 + 122n+1 is divisible by
(a) 113
(b) 123
(c) 133
(d) None of these
Show Answer & Explanation

Answer: (c) 133

Explanation:
1. For n = 1: 11³ + 12³ = 1331 + 1728 = 3059 = 133 × 23.
2. Induction step: 11k+3 + 122k+3 = 11(11k+2 + 122k+1) + 133 · 122k+1, which is a multiple of 133.

Teacher's Note:
1. Test n = 1 and factor the number you get.

Question 4: For a positive integer n, let \( a(n)=1+\frac12+\frac13+\frac14+\dots+\frac{1}{2^n-1} \). Then
(a) a(100) ≤ 100
(b) a(100) > 100
(c) a(200) ≤ 100
(d) a(200) < 100
Show Answer & Explanation

Answer: (a) a(100) ≤ 100

Explanation:
1. By induction, n/2 < a(n) ≤ n for every positive integer n.
2. So a(100) ≤ 100, while a(200) > 200/2 = 100.
3. Only option (a) is therefore true.

Teacher's Note:
1. Group the terms in blocks of 1, 2, 4, 8, ... terms and bound each block between 1/2 and 1.

Question 5: If m, n are any two odd positive integers with n < m, then the largest positive integer which divides all numbers of the type m² − n² is
(a) 4
(b) 6
(c) 8
(d) 9
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. Put m = 2k + 1, n = 2l + 1, so m² − n² = 4(k − l)(k + l + 1).
2. k − l and k + l + 1 have opposite parity, so their product is even and 8 divides m² − n².
3. 3² − 1² = 8, so no larger number works.

Teacher's Note:
1. m² − n² = (m − n)(m + n); both factors are even and one is a multiple of 4.

Question 6: Which one of the following is true for every positive integer n?
(a) \( \left(1+\frac1n\right)^n<n^2 \)
(b) \( \left(1+\frac1n\right)^n<2 \)
(c) \( \left(1+\frac1n\right)^n<n^3 \)
(d) \( \left(1+\frac1n\right)^n\ge 2 \)
Show Answer & Explanation

Answer: (d) \( \left(1+\frac1n\right)^n\ge 2 \)

Explanation:
1. For n = 1 the expression equals 2, so (b) fails and (a), (c) fail as well (2 is not less than 1).
2. By the binomial theorem, (1 + 1/n)n ≥ 1 + n · (1/n) = 2 for every n ∈ N.
3. So only (d) holds for all n.

Teacher's Note:
1. Check n = 1 first, then use Bernoulli's inequality.

Question 7: By the principle of induction, for all n ∈ N, 32n when divided by 8 leaves remainder
(a) 2
(b) 3
(c) 7
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. 32n = 9n, and 9 ≡ 1 (mod 8).
2. Hence 9n ≡ 1 (mod 8) for all n; the remainder is 1.

Teacher's Note:
1. Write 9 = 8 + 1.

Question 8: Let P(n): 2n < (1 × 2 × 3 × ... × n). Then the smallest positive integer for which P(n) is true is
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. P(1): 2 < 1, P(2): 4 < 2 and P(3): 8 < 6 are all false.
2. P(4): 16 < 24 is true, and the inequality persists for larger n by induction.

Teacher's Note:
1. Evaluate both sides for n = 1, 2, 3, 4.

Question 9: For 1 ≤ n ≤ 9, the number of values of n for which \( \dfrac{(n+2)!}{6(n-1)!} \) is divisible by n is
(a) 4
(b) 5
(c) 6
(d) 7
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. (n + 2)!/(6(n − 1)!) = n(n + 1)(n + 2)/6, so we need (n + 1)(n + 2)/6 to be an integer.
2. This holds for n = 1, 2, 4, 5, 7, 8 and fails for n = 3, 6, 9.
3. So there are 6 values.

Teacher's Note:
1. Cancel the common factor n and test n = 1, ..., 9.

Question 10: For every positive integer n, 7n − 3n is divisible by
(a) 7
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. n = 1 gives 4, n = 2 gives 40, so only 4 (of the options) can work.
2. Induction: 7k+1 − 3k+1 = 7(7k − 3k) + 4 · 3k, which is divisible by 4.

Teacher's Note:
1. Test n = 1 and n = 2 to eliminate options.

Question 11: If \( \dfrac{4^n}{n+1}<\dfrac{(2n)!}{(n!)^2} \), then P(n) is true for
(a) n ≥ 1
(b) n > 0
(c) n < 0
(d) n ≥ 2
Show Answer & Explanation

Answer: (d) n ≥ 2

Explanation:
1. n = 1: 2 < 2 is false; n = 2: 16/3 < 6 is true.
2. The inductive step works for every n ≥ 2, so the inequality holds exactly for n ≥ 2.

Teacher's Note:
1. Check n = 1 and n = 2 first.

Question 12: A student wants to prove a statement P(n) by induction. He proves that P(5) is true and that P(k + 1) is true whenever P(k) is true for any k ≥ 5. On this basis he can conclude that P(n) is true
(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5
Show Answer & Explanation

Answer: (c) for all n ≥ 5

Explanation:
1. P(5) is the base case and the inductive step takes P(k) to P(k + 1) for every k ≥ 5.
2. This gives P(5), P(6), P(7), ... so P(n) holds for all n ≥ 5. Nothing is known for n < 5.

Teacher's Note:
1. The base case fixes where the chain of implications starts.

Question 13: If n ∈ N and n > 1, then
(a) \( n!>\left(\frac{n+1}{2}\right)^n \)
(b) \( n!\ge\left(\frac{n+1}{2}\right)^n \)
(c) \( n!<\left(\frac{n+1}{2}\right)^n \)
(d) None of these
Show Answer & Explanation

Answer: (c) \( n!<\left(\frac{n+1}{2}\right)^n \)

Explanation:
1. n = 2: 2! = 2 and (3/2)² = 2.25, so n! < ((n+1)/2)n.
2. By AM–GM on 1, 2, ..., n: (n!)1/n < (1 + 2 + ... + n)/n = (n + 1)/2 for n > 1.

Teacher's Note:
1. Try n = 2, then think of AM–GM.

Question 14: For all n ∈ N, 41n − 14n is a multiple of
(a) 26
(b) 27
(c) 25
(d) None of these
Show Answer & Explanation

Answer: (b) 27

Explanation:
1. n = 1: 41 − 14 = 27; n = 2: 1681 − 196 = 1485 = 27 × 55.
2. In general an − bn is divisible by a − b = 27.

Teacher's Note:
1. a − b divides an − bn.

Question 15: For all n ∈ N, \( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dots+\dfrac{1}{1+2+3+\dots+n} \) is equal to
(a) \( \frac{3n}{n+1} \)
(b) \( \frac{n}{n+1} \)
(c) \( \frac{2n}{n-1} \)
(d) \( \frac{2n}{n+1} \)
Show Answer & Explanation

Answer: (d) \( \frac{2n}{n+1} \)

Explanation:
1. The r-th term is 2/(r(r + 1)) = 2(1/r − 1/(r + 1)) (r = 1 gives 1).
2. The sum telescopes to 2(1 − 1/(n + 1)) = 2n/(n + 1).

Teacher's Note:
1. 1 + 2 + ... + r = r(r + 1)/2.

Question 16: 10n + 3(4n+2) + 5 is divisible by (n ∈ N)
(a) 7
(b) 5
(c) 9
(d) 17
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. n = 1: 10 + 192 + 5 = 207 = 9 × 23, which is not divisible by 7, 5 or 17.
2. Induction: with S(k) = 10k + 3 · 4k+2 + 5, S(k+1) − S(k) = 9 · 10k + 9 · 4k+2 = 9(10k + 4k+2), so the difference is a multiple of 9.

Teacher's Note:
1. Compute n = 1 and factorise the result.

Question 17: The greatest positive integer which divides n(n + 1)(n + 2)(n + 3) for all n ∈ N is
(a) 2
(b) 6
(c) 24
(d) 120
Show Answer & Explanation

Answer: (c) 24

Explanation:
1. The product of four consecutive integers is divisible by 4! = 24.
2. For n = 1 it equals 24, so no larger number works.

Teacher's Note:
1. Evaluate at n = 1 and use the factorial divisibility of consecutive integers.

Question 18: Which of the following results is valid?
(a) (1 + x)n > (1 + nx) for all natural numbers n
(b) (1 + x)n ≥ (1 + nx) for all natural numbers n, where x > −1
(c) (1 + x)n ≤ (1 + nx) for all natural numbers n
(d) (1 + x)n < (1 + nx) for all natural numbers n
Show Answer & Explanation

Answer: (b) (1 + x)n ≥ (1 + nx) for all natural numbers n, where x > −1

Explanation:
1. This is Bernoulli's inequality, proved by induction: (1 + x)k+1 ≥ (1 + kx)(1 + x) = 1 + (k + 1)x + kx² ≥ 1 + (k + 1)x.
2. Equality occurs at n = 1, so the strict form (a) fails.

Teacher's Note:
1. Check n = 1 to rule out strict inequalities.

Question 19: When 2301 is divided by 5, the least positive remainder is
(a) 4
(b) 8
(c) 2
(d) 6
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. 24 = 16 ≡ 1 (mod 5).
2. 301 = 4 × 75 + 1, so 2301 ≡ 2 (mod 5). The remainder is 2.

Teacher's Note:
1. Find the cycle of powers of 2 modulo 5.

Question 20: Let S(k) = 1 + 3 + 5 + ... + (2k − 1) = 3 + k². Then which of the following is true?
(a) Principle of mathematical induction can be used to prove the formula
(b) S(k) ⇒ S(k + 1)
(c) S(k) ⇏ S(k + 1)
(d) S(1) is correct
Show Answer & Explanation

Answer: (b) S(k) ⇒ S(k + 1)

Explanation:
1. If S(k) holds, S(k + 1): 3 + k² + (2k + 1) = 3 + (k + 1)², which holds, so S(k) ⇒ S(k + 1).
2. But S(1): 1 = 4 is false, so the base case fails and the formula cannot be proved (the true sum is k²).

Teacher's Note:
1. An induction step alone does not prove a statement; check the base case too.

Question 21: For every positive integral value of n, 3n > n³ when
(a) n > 2
(b) n ≥ 3
(c) n ≥ 4
(d) n < 4
Show Answer & Explanation

Answer: (c) n ≥ 4

Explanation:
1. n = 3 gives 27 = 27, so the strict inequality fails for n = 3 (excluding (a), (b)).
2. n = 4: 81 > 64 and the step 3k+1 > 3k³ ≥ (k + 1)³ for k ≥ 4 proves it for all n ≥ 4.
3. Option (d) is false because n = 3 is included in it.

Teacher's Note:
1. Test n = 1, 2, 3, 4.

Question 22: If n is a natural number, then
(a) \( 1^2+2^2+\dots+n^2<\frac{n^3}{3} \)
(b) \( 1^2+2^2+\dots+n^2=\frac{n^3}{3} \)
(c) \( 1^2+2^2+\dots+n^2>n^3 \)
(d) \( 1^2+2^2+\dots+n^2>\frac{n^3}{3} \)
Show Answer & Explanation

Answer: (d) \( 1^2+2^2+\dots+n^2>\frac{n^3}{3} \)

Explanation:
1. 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6 = n³/3 + n²/2 + n/6.
2. This always exceeds n³/3, and it is less than n³ for n ≥ 2 (n = 1 gives equality), so (c) fails.

Teacher's Note:
1. Use the closed formula for the sum of squares.

Question 23: If P(n): 3n < n!, n ∈ N, then P(n) is true
(a) for n ≥ 6
(b) for n ≥ 7
(c) for n ≥ 3
(d) for all n
Show Answer & Explanation

Answer: (b) for n ≥ 7

Explanation:
1. 36 = 729 and 6! = 720, so P(6) is false.
2. 37 = 2187 < 5040 = 7!, and for k ≥ 7, 3k+1 = 3 · 3k < (k + 1) k!, so it is true for all n ≥ 7.

Teacher's Note:
1. Compute 3n and n! for n = 5, 6, 7.

Question 24: If P = n(n² − 1²)(n² − 2²)(n² − 3²) ... (n² − r²), n > r, n ∈ N, then P is necessarily divisible by
(a) (2r + 2)!
(b) (2r + 4)!
(c) (2r + 1)!
(d) None of these
Show Answer & Explanation

Answer: (c) (2r + 1)!

Explanation:
1. P = (n − r)(n − r + 1) ... (n − 1) n (n + 1) ... (n + r), a product of 2r + 1 consecutive integers.
2. A product of m consecutive integers is divisible by m!, so P is divisible by (2r + 1)!.

Teacher's Note:
1. Factorise each (n² − k²) as (n − k)(n + k).

Question 25: If 49n + 16n + k is divisible by 64 for every n ∈ N, then the greatest negative integral value of k is
(a) −1
(b) 1
(c) 2
(d) −2
Show Answer & Explanation

Answer: (a) −1

Explanation:
1. n = 1: 49 + 16 + k = 65 + k must be a multiple of 64, so k ≡ −1 (mod 64).
2. k = −1 works for all n: 49n + 16n − 1 = (1 + 48)n + 16n − 1, and the binomial expansion leaves 64 × integer.
3. The greatest negative value is −1.

Teacher's Note:
1. Test n = 1 to get k modulo 64.

Question 26: If \( \dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dots \) (n terms) \( =\dfrac{kn}{n+1} \), then k is equal to
(a) \( \frac14 \)
(b) \( \frac12 \)
(c) 1
(d) \( \frac18 \)
Show Answer & Explanation

Answer: (a) \( \frac14 \)

Explanation:
1. The r-th term is 1/(2r(2r + 2)) = (1/4)(1/r − 1/(r + 1)).
2. The sum telescopes to (1/4)(1 − 1/(n + 1)) = n/(4(n + 1)), so k = 1/4.

Teacher's Note:
1. Use partial fractions on 1/(r(r + 1)).

Question 27: For the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² + ..., let Sn be the sum of n terms. Then
(a) \( S_n=\frac{n(n+1)^2}{2} \), if n is even
(b) \( S_n=\frac{n^2(n+1)}{2} \), if n is odd
(c) Both (a) and (b) are true
(d) Both (a) and (b) are false
Show Answer & Explanation

Answer: (c) Both (a) and (b) are true

Explanation:
1. Check: S2 = 1 + 8 = 9 = 2 · 9/2, S4 = 50 = 4 · 25/2, S1 = 1 and S3 = 18 = 9 · 4/2.
2. Induction on both parities confirms both formulas, so (a) and (b) are true.

Teacher's Note:
1. Verify with n = 1, 2, 3, 4 and then prove by induction.

Question 28: The remainder when 54n is divided by 13 is
(a) 1
(b) 8
(c) 9
(d) 10
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. 54 = 625 = 13 × 48 + 1, so 54 ≡ 1 (mod 13).
2. Therefore 54n ≡ 1 (mod 13); the remainder is 1.

Teacher's Note:
1. Find 54 modulo 13.

Question 29: By mathematical induction, the numbers an are defined by a0 = 1, an+1 = 3n² + n + an (n ≥ 0). Then an is equal to
(a) n³ + n² + 1
(b) n³ − n² + 1
(c) n³ − n²
(d) n³ + n²
Show Answer & Explanation

Answer: (b) n³ − n² + 1

Explanation:
1. a1 = 1, a2 = 3 + 1 + 1 = 5, a3 = 12 + 2 + 5 = 19.
2. n³ − n² + 1 gives 1, 5, 19 for n = 1, 2, 3, matching; (n+1)³ − (n+1)² + 1 − (n³ − n² + 1) = 3n² + n, which proves the step.

Teacher's Note:
1. Compute the first few terms and test each option.

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Multiple Choice Questions (MCQs) for JEE Mathematics Mathematical Induction

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