JEE Mathematics Principles Of Mathematical Induction MCQs Set 10

Answer: (a) \( n>2 \)

Explanation:
1. Test \( n = 1 \): \( 1! = 1 \) and \( 2^0 = 1 \), so the inequality fails.
2. \( n = 2 \): \( 2! = 2 \) and \( 2^1 = 2 \), again not strict, so it fails.
3. \( n = 3 \): \( 6 \gt 4 \) holds, and if \( k! \gt 2^{k-1} \) for some \( k \ge 3 \) then \( (k+1)! = (k+1)k! \gt 2 \cdot 2^{k-1} = 2^k \).
4. So it is true exactly for \( n \gt 2 \).

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (c) true for all \( n\in N \)

Explanation:
1. For \( n = 1 \): left side = 1 and right side = \( 2! - 1 = 1 \), so \( P(1) \) is true.
2. Assume \( P(k) \). Adding \( (k+1)(k+1)! \) gives \( (k+1)! - 1 + (k+1)(k+1)! = (k+1)!(k+2) - 1 = (k+2)! - 1 \).
3. So \( P(k+1) \) is true and the statement holds for every natural number \( n \).

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (c) 133

Explanation:
1. For \( n = 1 \): \( 11^3 + 12^3 = 1331 + 1728 = 3059 = 133 x 23 \).
2. Step: \( 11^{k+3} + 12^{2k+3} = 11 \cdot 11^{k+2} + 144 \cdot 12^{2k+1} = 11(11^{k+2} + 12^{2k+1}) + 133 \cdot 12^{2k+1} \).
3. Both terms are divisible by 133, so the answer is 133.

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (a) \( a(100)\le 100 \)

Explanation:
1. Group the terms: 1, then 2 terms each at most 1/2, then 4 terms each at most 1/4, and so on. Each group contributes at most 1, and there are \( n \) groups.
2. So \( a(n) \le n \), which gives \( a(100) \le 100 \).
3. For \( a(200) \): the sum is about \( 200 \ln 2 + 0.58 \), roughly 139, which is more than 100. So \( (c) \) and \( (d) \) are false, and \( (b) \) is false.

Teacher's Note:
1. Harmonic sums grow like \( \ln \) of the number of terms; with \( 2^n - 1 \) terms this is about \( 0.69 n \).
2. Bounding by groups of powers of 2 is the standard induction-style argument.

Answer: (c) 8

Explanation:
1. Write \( m^2 - n^2 = (m-n)(m+n) \). Both factors are even, and their sum is \( 2m \), which is 2 times an odd number, so exactly one of them is a multiple of 4. The product is therefore a multiple of 8.
2. \( m = 3, n = 1 \) gives 8 and \( m = 5, n = 3 \) gives 16, so no larger common divisor exists.
3. \( m = 3, n = 1 \) gives 8, so the greatest common divisor cannot exceed 8. The answer is 8.

Teacher's Note:
1. Square of an odd number is always 1 more than a multiple of 8, so the difference of two odd squares is a multiple of 8.
2. Always test with the smallest examples to rule out larger candidates.

Answer: (d) 1

Explanation:
1. \( 3^{2n} = 9^n \), and \( 9 = 8 + 1 \).
2. So \( 9^n \) is congruent to 1 modulo 8 for every \( n \).
3. Check with \( n = 1 \): \( 9 = 8 + 1 \), remainder 1.

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (d) 4

Explanation:
1. \( n = 1 \): \( 2 \lt 1 \) is false. \( n = 2 \): \( 4 \lt 2 \) false. \( n = 3 \): \( 8 \lt 6 \) false.
2. \( n = 4 \): \( 16 \lt 24 \) is true.
3. If \( 2^k \lt k! \) for \( k \ge 4 \), then \( 2^{k+1} = 2 \cdot 2^k \lt (k+1) k! \) because \( k+1 \gt 2 \). So it stays true for all \( n \ge 4 \).

Teacher's Note:
1. For inequalities with factorials and exponentials, the factorial eventually wins; find the first \( n \) by direct trial.
2. Then use induction from that base value.

Answer: (c) 4

Explanation:
1. \( n = 1 \) gives \( 4, n = 2 \) gives \( 40, n = 3 \) gives 316.
2. gcd(4, 40, 316) = 4, so 7 and 5 and 3 fail for \( n = 1 \).
3. Also 7 is 3 modulo 4, so \( 7^n - 3^n \) is 0 modulo 4 for all \( n \). The answer is 4.

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (d) \( n\ge2 \)

Explanation:
1. \( n = 1 \): \( 4/2 = 2 \) and \( (2)!/(1)^2 = 2 \), so the strict inequality is false.
2. \( n = 2 \): 16/3 is about 5.33, and \( 4!/(2!)^2 = 6 \), so it is true.
3. The induction step multiplies the right side by \( (2k+2)(2k+1)/(k+1)^2 = 2(2k+1)/(k+1) \) and the left by \( 4(k+1)/(k+2) \); the former is larger for \( k \ge 2 \). So the inequality holds for \( n \ge 2 \).

Teacher's Note:
1. Always check the first few values before trusting the inequality sign.
2. Here the equality at \( n = 1 \) is why the statement starts only from \( n = 2 \).

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

Explanation:
1. \( n = 2 \): \( 2! = 2 \) and \( (3/2)^2 = 2.25 \), so \( n! \) is smaller.
2. In general pair the factors \( k \) and \( n+1-k \); each pair has product at most \( ((n+1)/2)^2 \) by AM-GM, with strict inequality somewhere for \( n \gt 1 \).
3. Hence \( n! \lt ((n+1)/2)^n \).

Teacher's Note:
1. AM-GM on the pairs \( k(n+1-k) \) is the slickest route; induction also works.
2. For \( n = 1 \) both sides are equal, which is why the question says \( n \gt 1 \).

Answer: (b) 27

Explanation:
1. \( a^n - b^n \) is divisible by \( a - b \).
2. Here \( a - b = 41 - 14 = 27 \).
3. Check \( n = 1 \): 27, and \( n = 2 \): \( 1681 - 196 = 1485 = 27 x 55 \).

Teacher's Note:
1. The factor theorem form: \( x - y \) divides \( x^n - y^n \).
2. Induction proof: \( 41^{k+1} - 14^{k+1} = 41(41^k - 14^k) + 27 \cdot 14^k \).

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

Explanation:
1. The k-th term is \( 1/(1+2+...+k) = 2/(k(k+1)) = 2(1/k - 1/(k+1)) \) for \( k = 1 \) to \( n \), with the first term 1 equal to 2/(1*2).
2. The sum telescopes to \( 2(1 - 1/(n+1)) = 2n/(n+1) \).
3. Check \( n = 1 \): \( 2/2 = 1 \), which matches the first term.

Teacher's Note:
1. Telescoping is usually quicker than induction for such sums, but induction verifies the formula.
2. Check the formula with \( n = 2 \): \( 1 + 1/3 = 4/3 = 4/3 \).

Answer: (c) 9

Explanation:
1. \( n = 1 \): \( 10 + 192 + 5 = 207 = 9 x 23 \).
2. \( n = 2 \): \( 100 + 768 + 5 = 873 = 9 x 97 \).
3. Step: \( 10^{k+1} + 3\cdot 4^{k+3} + 5 = 10(10^k + 3\cdot 4^{k+2} + 5) - 27\cdot 4^{k+2} - 45 \), and both extra terms are multiples of 9. Divisible by 9; 207 is not divisible by 7, 5 or 17.

Teacher's Note:
1. Induction needs a base case and a step showing \( P(k) \) implies \( P(k+1) \); check both whenever a statement is claimed for all \( n \).
2. For divisibility questions, test small values of \( n \) first to narrow the options, then confirm by the induction step.

Answer: (c) 24

Explanation:
1. Product of 4 consecutive integers is divisible by \( 4! = 24 \).
2. \( n = 1 \) gives \( 1 x 2 x 3 x 4 = 24 \), so nothing larger than 24 can divide every value.

Teacher's Note:
1. The product of \( r \) consecutive integers is always divisible by \( r! \).
2. The first value \( n = 1 \) shows the bound is tight.

Answer: (b) \( (1+x)^n\ge(1+nx) \) for all natural numbers n, where \( x>-1 \)

Explanation:
1. This is Bernoulli's inequality: \( (1+x)^n \ge 1 \) + nx for \( x \gt -1 \).
2. For \( n = 1 \) equality holds, so the strict forms \( (a) \) and \( (d) \) fail.
3. Induction step: \( (1+x)^{k+1} \) >= (1+kx)(1+x) = \( 1 + (k+1)x + kx^2 \ge 1 + (k+1)x \).

Teacher's Note:
1. Bernoulli's inequality needs \( x \gt -1 \) so that \( 1 + x \) is positive when multiplying through.
2. It is a frequent tool for proving limits and AM-GM type results.

Answer: (c) 2

Explanation:
1. \( 2^4 = 16 \), which leaves remainder 1 on division by 5.
2. \( 301 = 4 x 75 + 1 \), so \( 2^{301} = (2^4)^{75} \cdot 2 \), which leaves remainder 2.

Teacher's Note:
1. Powers repeat modulo a number; find the cycle length first.
2. The cycle of \( 2^n \) mod 5 is 2, 4, 3, 1, of length 4.

Answer: (b) \( S(k)\Rightarrow S(k+1) \)

Explanation:
1. Assume \( S(k) \) holds, so the sum equals \( 3 + k^2 \).
2. Adding the next term \( 2k + 1 \): \( 3 + k^2 + 2k + 1 = 3 + (k+1)^2 \), which is exactly \( S(k+1) \).
3. For \( k \ge 4, 3^{k+1} = 3 \cdot 3^k \gt 3k^3 \ge (k+1)^3 \), so it holds for all \( n \ge 4 \). It fails at \( n = 3 \), hence \( n \ge 4 \).

Teacher's Note:
1. The inequality holds for \( n = 1, 2 \) and then from \( n = 4 \) onwards; \( n = 3 \) is the only failure.
2. So the correct range among the options is \( n \ge 4 \).

Answer: (c) \( n\ge4 \)

Explanation:
1. \( n = 3 \) gives \( 27 = 27 \), so the strict inequality fails; \( n = 1, 2 \) hold \( (3 \gt 1, 9 \gt 8) \).
2. \( n = 4 \): \( 81 \gt 64 \) is true.
3. For \( k \ge 4, 3^{k+1} = 3 \cdot 3^k \gt 3k^3 \ge (k+1)^3 \). So it holds for all \( n \ge 4 \) and fails at \( n = 3 \), so the answer is \( n \ge 4 \) as the stated continuous range.

Teacher's Note:
1. The range must hold for every \( n \) in it; \( n = 3 \) is the single failure.
2. So the right option is the one starting at 4.

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

Explanation:
1. The sum equals \( n(n+1)(2n+1)/6 = (2n^3 + 3n^2 + n)/6 \).
2. This exceeds \( n^3/3 = 2n^3/6 \) because \( 3n^2 + n \gt 0 \).
3. For \( n = 1 \): \( 1 \gt 1/3 \) confirms it.

Teacher's Note:
1. Compare with the integral of \( x^2 \), which is \( n^3/3 \).
2. The sum always exceeds it for natural \( n \).

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

Explanation:
1. Each factor \( n^2 - k^2 = (n-k)(n+k) \), so \( P = (n-r)(n-r+1)...(n-1)n(n+1)...(n+r) \).
2. That is a product of \( 2r + 1 \) consecutive integers.
3. A product of \( m \) consecutive integers is divisible by \( m! \), so \( P \) is divisible by \( (2r+1)! \).

Teacher's Note:
1. Product of \( m \) consecutive integers equals \( m! \) times a binomial coefficient.
2. Here \( m = 2r + 1 \).

Answer: (a) \( \frac14 \)

Explanation:
1. The r-th term is \( 1/(2r(2r+2)) = 1/(4r(r+1)) = (1/4)(1/r - 1/(r+1)) \).
2. The sum telescopes to \( (1/4)(1 - 1/(n+1)) = n/(4(n+1)) \).
3. So \( k = 1/4 \).

Teacher's Note:
1. Check with \( n = 1 \): \( 1/8 = k/2 \) gives \( k = 1/4 \).
2. Partial fractions turn such sums into telescoping ones.

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

Explanation:
1. Check small cases: \( S_1 = 1 \) and the odd formula gives \( 1 \cdot 2 / 2 = 1 \); \( S_2 = 1 + 8 = 9 \) and the even formula gives \( 2 \cdot 9 / 2 = 9 \).
2. \( S_3 = 9 + 9 = 18 \) and the odd formula gives \( 9 \cdot 4 / 2 = 18 \).
3. Both formulas hold in general by induction (add one term to go from even to odd and vice versa), so both are true.

Teacher's Note:
1. Series with alternating patterns need separate formulas for even and odd \( n \).
2. Verify small values first, then use induction in two steps.

Answer: (a) 1

Explanation:
1. \( 5^2 = 25 \), which is \( -1 \) modulo 13.
2. So \( 5^4 = 625 = 1 \) mod 13, and \( 5^{4n} = (5^4)^n \) is 1 mod 13.

Teacher's Note:
1. Find the smallest power giving 1 modulo the divisor.
2. \( 625 = 13 x 48 + 1 \).

Answer: (b) \( n^3-n^2+1 \)

Explanation:
1. \( a_1 = 3(0) + 0 + 1 = 1, a_2 = 3 + 1 + 1 = 5, a_3 = 12 + 2 + 5 = 19 \).
2. Test option \( (b) \): \( n = 1 \) gives \( 1, n = 2 \) gives \( 8 - 4 + 1 = 5, n = 3 \) gives \( 27 - 9 + 1 = 19 \).
3. Induction: \( (n+1)^3 - (n+1)^2 + 1 - (n^3 - n^2 + 1) = 3n^2 + n \), which matches the recurrence.

Teacher's Note:
1. Compute the first few terms and compare with options.
2. Then prove by induction using the difference of consecutive terms.