Practice MCQs for JEE Mathematics Mathematical Induction
Access targeted multiple-choice questions for Mathematical Induction designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
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Access the complete set of multiple-choice questions for Mathematical Induction below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) all natural numbers n
(b) nothing can be said
(c) n > 1
(d) n > 2
Show Answer & Explanation
Answer: (b) nothing can be said
Explanation:
1. The inductive step holds, but induction also needs a true base case.
2. \( P(1) \) would require \( 2 = 1\cdot 2 + 2 = 4 \), which is false, so no value of \( n \) can be used to start the chain: nothing can be concluded.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 2
(b) 6
(c) 24
(d) 120
Show Answer & Explanation
Answer: (c) 24
Explanation:
1. The product of four consecutive integers contains a multiple of \( 4, a \) multiple of 2 and a multiple of 3, hence is divisible by 24.
2. At \( n = 1 \) the product is 24, so no larger integer works.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(2\sin\!\left(\dfrac\pi{2^{n+1}}\right)\)
(b) \(\sqrt2\cos(2^{n+1}\pi)\)
(c) \(2\cos\!\left(\dfrac\pi{2^{n+1}}\right)\)
(d) none of these
Show Answer & Explanation
Answer: (c) \(2\cos\!\left(\dfrac\pi{2^{n+1}}\right)\)
Explanation:
1. \( n = 1 \): \(\sqrt2=2\cos\tfrac\pi4\), true.
2. If the value for \( n \) is \(2\cos\tfrac\pi{2^{k+1}}\), then for \( n = k+1 \) it is \(\sqrt{2+2\cos\tfrac\pi{2^{k+1}}}=2\cos\tfrac\pi{2^{k+2}}\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(a(100)\le100\)
(b) \(a(100)>100\)
(c) \(a(200)\le100\)
(d) \(a(200)<100\)
Show Answer & Explanation
Answer: (a) \(a(100)\le100\)
Explanation:
1. Group the terms: the j-th group has \(2^j\) terms, each at most \(2^{-j}\), so each group is at most 1 and \(a(n)\le n\).
2. Hence \(a(100)\le100\). Similarly each group exceeds 1/2, so \(a(n)>n/2\) and \(a(200)>100\), which rules out the other options.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) even values of n only
(b) odd values of n only
(c) all values of n
(d) no value of n
Show Answer & Explanation
Answer: (c) all values of n
Explanation:
1. \( n = 1 \): both sides equal 1. \( n = 2 \): \( 1/2 + 1/3 = 5/6 = 1 - 1/2 + 1/3 \). \( n = 3 \): both sides equal 47/60.
2. The identity is proved by induction for every natural number \( n \) (it is the classical identity for the alternating harmonic sum).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(\dfrac1{n+1}\)
(b) \(\dfrac n{n+1}\)
(c) \(\dfrac1{n(n+1)}\)
(d) none of these
Show Answer & Explanation
Answer: (b) \(\dfrac n{n+1}\)
Explanation:
1. \(\dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1}\), so the sum telescopes to \(1-\dfrac1{n+1}=\dfrac n{n+1}\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \((n-1)^2\)
(b) \(n(n+1)\)
(c) \((n+1)^2\)
(d) none of these
Show Answer & Explanation
Answer: (c) \((n+1)^2\)
Explanation:
1. \(1+\dfrac{2k+1}{k^2}=\dfrac{(k+1)^2}{k^2}\).
2. The product telescopes to \(\dfrac{(n+1)^2}{1^2}=(n+1)^2\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(n\ge5\)
(b) \(n>2\)
(c) \(n>3\)
(d) \(n<5\)
Show Answer & Explanation
Answer: (a) \(n\ge5\)
Explanation:
1. \( n = 2 \): 4 > 4 false; \( n = 3 \): 8 > 9 false; \( n = 4 \): 16 > 16 false; \( n = 5 \): 32 > 25 true.
2. If \(2^k>k^2\) with \(k\ge5\), then \(2^{k+1}>2k^2\ge(k+1)^2\) since \(k^2\ge2k+1\). So it holds for every \(n\ge5\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(n\ge1\)
(b) \(n\ge2\)
(c) \(n>0 \text{ only}\)
(d) \(n<0\)
Show Answer & Explanation
Answer: (b) \(n\ge2\)
Explanation:
1. \( n = 1 \): \(\tfrac42=2\) versus \(\tfrac{2}{1}=2\): not strictly less, so false.
2. \( n = 2 \): \(\tfrac{16}3\approx5.33<6\), true, and induction extends it to all \(n\ge2\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(n\ge6\)
(b) \(n\ge3\)
(c) \(n\ge7\)
(d) all n
Show Answer & Explanation
Answer: (c) \(n\ge7\)
Explanation:
1. \(3^6=729>720=6!\) so \( n = 6 \) fails; \(3^7=2187<5040=7!\).
2. If \(3^k<k!\) for \(k\ge7\), then \(3^{k+1}<3\cdot k!<(k+1)\,k!\). So it holds for all \(n\ge7\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 2
(b) 7
(c) 27
(d) 11
Show Answer & Explanation
Answer: (d) 11
Explanation:
1. \( n = 1 \): \( 2\cdot 64 + 81 = 209 = 11\cdot 19 \).
2. Modulo 11: \(2\cdot4^{2n+1}=8\cdot16^n\equiv8\cdot5^n\) and \(3^{3n+1}=3\cdot27^n\equiv3\cdot5^n\), so the sum is \(11\cdot5^n\equiv0\). The gcd for \( n \) up to 30 is exactly 11.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 6
(b) 4
(c) 10
(d) none of these
Show Answer & Explanation
Answer: (a) 6
Explanation:
1. \(n(n^2-1)=(n-1)n(n+1)\), a product of three consecutive integers.
2. It always contains a multiple of 2 and a multiple of 3, so it is divisible by 6 (and \( n = 2 \) gives 6, so 4 and 10 fail).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) -2
(b) -3
(c) -4
(d) -1
Show Answer & Explanation
Answer: (d) -1
Explanation:
1. \( n = 1 \): \(49+16+\lambda=65+\lambda\) must be a multiple of 64, so \(\lambda=-1\) (or \( -65, ... \)).
2. \(49^n=(1+48)^n=1+48n+64(\ldots)\), so \(49^n+16n-1=64n+64(\ldots)\) is divisible by 64 for all \( n \).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(\dfrac{n(n+1)}{4(n+2)(n+3)}\)
(b) \(\dfrac{n(n+2)}{4(n+1)(n+3)}\)
(c) \(\dfrac{n(n+3)}{4(n+1)(n+2)}\)
(d) none of these
Show Answer & Explanation
Answer: (c) \(\dfrac{n(n+3)}{4(n+1)(n+2)}\)
Explanation:
1. \(\dfrac1{k(k+1)(k+2)}=\dfrac12\left[\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}\right]\).
2. The sum telescopes: \(\dfrac12\left[\dfrac12-\dfrac1{(n+1)(n+2)}\right]=\dfrac{n(n+3)}{4(n+1)(n+2)}\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 17
(b) 19
(c) 23
(d) 25
Show Answer & Explanation
Answer: (a) 17
Explanation:
1. \( n = 1 \): \( 3\cdot 125 + 16 = 391 = 17\cdot 23 \). \( n = 2 \): \( 3\cdot 3125 + 128 = 9503 = 17\cdot 559 \).
2. \(3\cdot5^{2n+1}+2^{3n+1}=15\cdot25^n+2\cdot8^n\) and \(25\equiv8\pmod{17}\), so it is congruent to \(17\cdot8^n\equiv0\). The gcd for \( n \) up to 29 is exactly 17.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(\dfrac{(2n+1)3^{n+1}+3}4\)
(b) \(\dfrac{(2n-1)3^{n+1}+3}4\)
(c) \(\dfrac{(2n+1)3^n+3}4\)
(d) \(\dfrac{(2n-1)3^{n+1}+1}4\)
Show Answer & Explanation
Answer: (b) \(\dfrac{(2n-1)3^{n+1}+3}4\)
Explanation:
1. \( n = 1 \): \(\dfrac{1\cdot9+3}4=3\) correct.
2. If it holds for \( k \), adding \((k+1)3^{k+1}\): \(\dfrac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}4=\dfrac{(6k+3)3^{k+1}+3}4=\dfrac{(2k+1)3^{k+2}+3}4\), the formula for \( k+1 \).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) \(\dfrac{3n}{n+1}\)
(b) \(\dfrac{2n}{n+1}\)
(c) \(\dfrac n{n+1}\)
(d) \(\dfrac{2n}{n-1}\)
Show Answer & Explanation
Answer: (b) \(\dfrac{2n}{n+1}\)
Explanation:
1. \(\dfrac1{1+2+\cdots+k}=\dfrac2{k(k+1)}=2\left(\dfrac1k-\dfrac1{k+1}\right)\) (also true for \( k = 1 \)).
2. The sum is \(2\left(1-\dfrac1{n+1}\right)=\dfrac{2n}{n+1}\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation
Answer: (c) 2
Explanation:
1. \(2^4=16\equiv1\pmod5\) and \(301=4\cdot75+1\).
2. \(2^{301}\equiv2\pmod5\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 2
(b) 3
(c) 7
(d) 1
Show Answer & Explanation
Answer: (d) 1
Explanation:
1. \(3^2=9\equiv1\pmod8\), so \(3^{2n}=9^n\equiv1\).
2. Induction: \( 9^{k+1} = 9\cdot 9^k = 8\cdot 9^k + 9^k \), and \( 9^k \) leaves remainder 1.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 133
(b) 123
(c) 143
(d) 11
Show Answer & Explanation
Answer: (a) 133
Explanation:
1. \(11^{n+2}+12^{2n+1}=121\cdot11^n+12\cdot144^n\), and \(144\equiv11\pmod{133}\) so it equals \(11^n(121+12)\equiv0\) modulo 133.
2. \( n = 1 \): \(1331+1728=3059=133\cdot23\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 26
(b) 41
(c) 27
(d) 55
Show Answer & Explanation
Answer: (c) 27
Explanation:
1. \(a^n-b^n\) is divisible by \(a-b\), here \(41-14=27\).
2. \( n = 1 \) gives exactly 27, so 27 is the largest such constant.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 4
(b) 8
(c) 16
(d) 2
Show Answer & Explanation
Answer: (b) 8
Explanation:
1. \(m^2-n^2=(m-n)(m+n)\); both factors are even and one of them is divisible by 4, so 8 divides it.
2. \(3^2-1^2=8\), so no larger number works.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 16
(b) 8
(c) 32
(d) 64
Show Answer & Explanation
Answer: (d) 64
Explanation:
1. \( n = 1 \): \(81-8-9=64\). \( n = 2 \): \(729-16-9=704=64\cdot11\).
2. \(9^{n+1}=(1+8)^{n+1}=1+8(n+1)+64(\ldots)\), so the expression is divisible by 64, and since \( n = 1 \) gives 64 it is the greatest.
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
(a) 8
(b) 1
(c) 5
(d) 12
Show Answer & Explanation
Answer: (a) 8
Explanation:
1. \(5^4=625=48\cdot13+1\equiv1\).
2. \(5^{99}=5^{96}\cdot5^3\equiv125\equiv125-117=8\pmod{13}\).
Teacher's Note:
1. Check the base case first and then show that \( P(k) \) implies \( P(k+1) \); both steps are needed.
2. For divisibility questions, test small values of \( n \) to find the number and then confirm with the inductive step.
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Multiple Choice Questions (MCQs) for JEE Mathematics Mathematical Induction
JEE Mathematics Mathematical Induction Objective Test Questions
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