JEE Mathematics Principles Of Mathematical Induction MCQs Set 09

Answer: (c) all \( n>3 \)

Explanation:
1. n = 1, 2, 3: \( 2<1 \) false, \( 4<2 \) false, \( 8<6 \) false.
2. n = 4: \( 16<24 \) is true. If \( 2^k<k! \) with \( k\ge4 \) then \( 2^{k+1}<2\cdot k!<(k+1)\cdot k! \), so P(n) holds for all \( n>3 \).

Teacher's Note:
1. Find the first n for which the base case works, then check the induction step.

Answer: (d) nothing can be said

Explanation:
1. The induction step is valid: \( k(k+1)+2+2(k+1)=(k+1)(k+2)+2 \).
2. But the base case fails: \( P(1) \) says \( 2=1\cdot2+2=4 \). Without a true starting case, induction proves nothing, so nothing can be said.

Teacher's Note:
1. Induction needs both a true base case and a valid step.

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

Explanation:
1. Adding \( 2k+1 \) to both sides of \( T(k) \) gives \( k^2+2k+1+10=(k+1)^2+10 \), so \( T(k)\Rightarrow T(k+1) \).
2. \( T(1) \) says \( 1=11 \), which is false, so T(n) is not true for all n.

Teacher's Note:
1. A valid step does not make the statement true without a base case.

Answer: (d) 4

Explanation:
1. \( n=1,2,3 \): \( 2<1 \), \( 4<2 \), \( 8<6 \) are all false.
2. \( n=4 \): \( 16<24 \) is true.

Teacher's Note:
1. Test small values of n in order.

Answer: (a) 11

Explanation:
1. For n = 1: \( 10+1=11 \).
2. \( 10^{2k+1}+1=100\cdot10^{2k-1}+1=100(10^{2k-1}+1)-99 \), a multiple of 11. So it is divisible by 11.

Teacher's Note:
1. The base case gives the divisor.

Answer: (d) \( \dfrac{n(n+1)(2n+1)}{6} \)

Explanation:
1. For n = 1: \( \frac{1\cdot2\cdot3}{6}=1 \).
2. If true for k, then adding \( (k+1)^2 \) gives \( \frac{(k+1)(k+2)(2k+3)}{6} \), which is the formula for \( k+1 \).

Teacher's Note:
1. Check n = 1 against each option.

Answer: (a) 24

Explanation:
1. P(1) = 14 + 15 - 5 = 24, P(2) = 168 = 24 x 7.
2. \( P(k+1)-5P(k)=4\cdot7^k+20=4(7^k+5) \), and \( 7^k+5 \) is a multiple of 6 because \( 7^k\equiv1\pmod6 \). So the difference is a multiple of 24, and divisibility by 24 carries over.

Teacher's Note:
1. The gcd of the first two values bounds the answer.

Answer: (b) \( \dfrac{n}{n+1} \)

Explanation:
1. Each term is \( \frac1k-\frac1{k+1} \), so the sum telescopes to \( 1-\frac1{n+1} \).
2. That is \( \frac{n}{n+1} \) (n = 1 gives \( \frac12 \)).

Teacher's Note:
1. Use partial fractions.

Answer: (c) 4

Explanation:
1. n = 1: 4; n = 2: 40; n = 3: 316 - all divisible by 4.
2. \( 7^{k+1}-3^{k+1}=7(7^k-3^k)+4\cdot3^k \), so divisibility by 4 carries over.

Teacher's Note:
1. Write \( 7^{k+1}-3^{k+1} \) in terms of \( 7^k-3^k \).

Answer: (b) \( \dfrac{n(n+3)}{4(n+1)(n+2)} \)

Explanation:
1. For n = 1: \( \frac{1\cdot4}{4\cdot2\cdot3}=\frac16 \), which matches the first term.
2. Only option (b) gives \( \frac16 \) at n = 1 and also \( \frac16+\frac1{24}=\frac{5}{24} \) at n = 2 (\( \frac{2\cdot5}{4\cdot3\cdot4}=\frac5{24} \)).

Teacher's Note:
1. Check n = 1 and n = 2 against each option.

Answer: (c) 133

Explanation:
1. n = 1: \( 1331+1728=3059=133\times23 \).
2. \( 11^{k+3}+12^{2k+3}=11(11^{k+2}+12^{2k+1})+133\cdot12^{2k+1} \), so divisibility by 133 carries over.

Teacher's Note:
1. Write the (k+1) term using the k term.

Answer: (b) 27

Explanation:
1. n = 1: 27.
2. \( 41^{k+1}-14^{k+1}=41(41^k-14^k)+27\cdot14^k \), so divisibility by 27 carries over.

Teacher's Note:
1. The base case gives 27.

Answer: (a) 1

Explanation:
1. \( 5^4=625=48\times13+1 \), so \( 5^4\equiv1\pmod{13} \).
2. Hence \( 5^{4n}\equiv1 \), and the remainder is 1.

Teacher's Note:
1. 625 leaves remainder 1 on division by 13.

Answer: (c) 8

Explanation:
1. \( m^2-n^2=(m-n)(m+n) \) with both factors even, and one of them a multiple of 4 (their sum \( 2m \) is not a multiple of 4 when m is odd), so the product is a multiple of 8.
2. Examples: \( 9-1=8 \), \( 25-9=16 \), \( 25-1=24 \); the gcd is 8.

Teacher's Note:
1. Odd squares leave remainder 1 modulo 8.

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

Explanation:
1. n = 2: \( 4\cdot1<4 \) is false, so (c) fails.
2. n = 3: \( 8\cdot2=16<27 \) is true, and the step \( 2^{k+1}k!<(k+1)^{k+1} \) follows from \( \left(1+\frac1k\right)^k\ge2 \). So the inequality holds for \( n>2 \).

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

Answer: (b) 17

Explanation:
1. n = 1: \( 375+16=391=17\times23 \); n = 2: \( 9375+128=9503=17\times559 \).
2. 23 does not divide 9503, so the common divisor is 17.

Teacher's Note:
1. Compute n = 1 and n = 2 and take the common factor.

Answer: (b) \( \dfrac{(2n-1)3^{n+1}+3}{4} \)

Explanation:
1. For n = 1 the sum is 3, and \( \frac{(2-1)\cdot9+3}{4}=3 \).
2. Option (a) gives 15/4 at n = 1, (c) gives 9/4, (d) gives 10/4, so only (b) fits; the step follows by adding \( (k+1)3^{k+1} \).

Teacher's Note:
1. Substitute n = 1 into each option.

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

Explanation:
1. The k-th term is \( \frac{2}{k(k+1)}=2\left(\frac1k-\frac1{k+1}\right) \).
2. The sum telescopes to \( 2\left(1-\frac1{n+1}\right)=\frac{2n}{n+1} \).

Teacher's Note:
1. The sum of the first k natural numbers is \( k(k+1)/2 \).

Answer: (c) 9

Explanation:
1. n = 1: \( 10+192+5=207=9\times23 \); n = 2: \( 100+768+5=873=9\times97 \).
2. Only 9 divides both.

Teacher's Note:
1. Compute two values and take the common divisor.

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

Explanation:
1. n = 1: \( 1=2!-1 \).
2. If true for k, adding \( (k+1)(k+1)! \) gives \( (k+1)!(1+k+1)-1=(k+2)!-1 \).

Teacher's Note:
1. Add \( (k+1)(k+1)! \) to both sides.

Answer: (b) \( n\ge1 \)

Explanation:
1. n = 1: \( 1\ge1 \); n = 2: \( \frac94\ge2 \).
2. By the AM-GM inequality on \( 1,2,\dots,n \) the mean \( \frac{n+1}{2} \) is at least \( (n!)^{1/n} \) for every \( n\ge1 \).

Teacher's Note:
1. Equality holds at n = 1, so the inequality holds for all n.

Answer: (d) \( n\ge3 \)

Explanation:
1. n = 2: \( 4>4 \) is false.
2. n = 3: \( 36>27 \) is true, and the inequality is maintained for larger n. So it holds for \( n\ge3 \).

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