JEE Mathematics Principles Of Mathematical Induction MCQs Set 01

Answer: (b) N, the set of natural numbers

Explanation:
1. The principle of mathematical induction starts at n = 1 and moves from n = k to n = k + 1.
2. So it applies to statements indexed by the natural numbers N.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

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

Explanation:
1. For n = 1: LHS = 1, RHS = 2! - 1 = 1. For n = 2: LHS = 1 + 4 = 5, RHS = 3! - 1 = 5.
2. If it holds for n = k, then adding \( (k+1)(k+1)! \) gives \( (k+1)! - 1 + (k+1)(k+1)! = (k+2)! - 1 \), so it holds for n = k + 1. It is true for all natural numbers.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) 2

Explanation:
1. n = 1: \( 1! = 1 \) and \( ((1+1)/2)^1 = 1 \), so 1 < 1 is false.
2. n = 2: \( 2! = 2 \) and \( (3/2)^2 = 2.25 \), so 2 < 2.25 is true. The smallest value is 2.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) \( S(K) \Rightarrow S(K+1) \)

Explanation:
1. If \( S(K) \) holds, adding \( 2K + 1 \) to both sides gives \( 1 + 3 + \dots + (2K+1) = K^2 + 3 + 2K + 1 = (K+1)^2 + 3 \), so \( S(K+1) \) holds.
2. But \( S(1) \) says \( 1 = 1 + 3 \), which is false, so the induction cannot start and the formula is not true (the true sum is \( K^2 \)).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) even

Explanation:
1. Of two consecutive numbers one is even, so the product is always even.
2. For n = 1 the product is 2, which is not a multiple of 3 or 4.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (a) no value of n

Explanation:
1. \( (n+1)^2 + (n+1) = n^2 + n + 2(n+1) \), so if \( n^2 + n \) is odd, so is the next one: \( P(n) \Rightarrow P(n+1) \).
2. But \( n(n+1) \) is always even, so P(1) is false, and the induction never starts. P(n) is true for no value of n.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) \( \{ 6k : k \in Z \} \)

Explanation:
1. n(n+1)(2n+1) is 6 times the sum of the first n squares, so it is a multiple of 6.
2. For n = 1 the value is 6 and for n = 2 it is 30, which are not multiples of 12, 18 or 24.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) \( a_n < 7 \) for all \( n \ge 1 \)

Explanation:
1. \( a_1 = \sqrt{7} < 7 \). If \( a_m < 7 \), then \( a_{m+1} = \sqrt{7 + a_m} < \sqrt{14} < 7 \).
2. So \( a_n < 7 \) for all n. Also \( a_1 = \sqrt 7 > 2 \), so \( a_n < 2 \) fails.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) \( \dfrac{\sqrt{a+nx} - \sqrt a}{x} \)

Explanation:
1. Rationalise each term: \( \dfrac{1}{\sqrt{a+(r-1)x} + \sqrt{a+rx}} = \dfrac{\sqrt{a+rx} - \sqrt{a+(r-1)x}}{x} \).
2. The terms telescope, leaving the last square root minus the first, divided by x.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (a) Natural number

Explanation:
1. The expression equals \( \dfrac{n(n+1)(n+2)(n+3)}{24} \), the number of ways of choosing 4 objects from n + 3, so it is a positive integer.
2. So it is a natural number.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) all odd \( n \)

Explanation:
1. \( a^n + b^n \) has the factor \( a + b \) when n is odd, since \( x + y \) divides \( x^n + y^n \) for odd n.
2. For n = 2, \( a^2 + b^2 \) is not divisible by \( a + b \) in general (for example a = 2, b = 4: 20 is not divisible by 6).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) natural number

Explanation:
1. By induction on n: \( (n+1)^p - (n+1) \equiv n^p + 1 - n - 1 = n^p - n \pmod p \), since the middle binomial coefficients are multiples of p.
2. So the statement holds for every natural number n (Fermat's little theorem).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) \( r! \)

Explanation:
1. The product of r consecutive natural numbers is divisible by r!.
2. For example for r = 3, the products 24, 60, 120, 210, ... have highest common factor 6 = 3!.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (a) for all \( n \ge 5 \)

Explanation:
1. The base case is n = 5, and the step carries the truth to every later natural number.
2. So P(n) is true for all n >= 5 (nothing is shown for n < 5).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) None of the above

Explanation:
1. The value of \( n(n+1) + 1 \) is 7 for n = 2, 13 for n = 3 and 21 for n = 4, all odd.
2. A product of consecutive numbers is even, and adding 1 makes it odd, so none of the statements is even.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) 7

Explanation:
1. For n = 2: 2^4 + 1 = 17, units digit 7.
2. For even n, 2^(2n) = 4^n with n even ends in 6 (4^2 = 16, 4^4 = 256, ...), so adding 1 gives units digit 7.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) all \( n > 2, n \in N \)

Explanation:
1. n = 1: 1 > 1 is false. n = 2: 4 > 4 is false. n = 3: 36 > 27 is true.
2. The induction step works from n = 3 onwards, so it holds for all \( n > 2 \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (a) \( n > 1, n \in N \)

Explanation:
1. n = 1: 1 < 1 is false. n = 2: 5/4 < 3/2 is true.
2. The step: \( \frac1{k+1} + \frac1{(k+1)^2} \)... works from n = 2 onward, so it is true for n > 1.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) 425

Explanation:
1. The odd cubes sum to 1 + 27 + 125 + 343 + 729 = 1225 and the even cubes to 8 + 64 + 216 + 512 = 800.
2. 1225 - 800 = 425.

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) \( \dfrac{n}{6n+4} \)

Explanation:
1. The r-th term is \( \frac1{(3r-1)(3r+2)} = \frac13\left(\frac1{3r-1} - \frac1{3r+2}\right) \), which telescopes.
2. Sum \( = \frac13\left(\frac12 - \frac1{3n+2}\right) = \frac{n}{2(3n+2)} = \frac{n}{6n+4} \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) \( \dfrac{n(2n^2 + 9n + 1)}{6} \)

Explanation:
1. The r-th term is \( r^2 + 2r - 1 \).
2. Sum \( = \frac{n(n+1)(2n+1)}{6} + n(n+1) - n = \frac{n(2n^2 + 9n + 1)}{6} \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

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

Explanation:
1. Sum \( = \frac12\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)(n+2)}{6} \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (d) \( 3n(n+1)(n+2) \)

Explanation:
1. The r-th term is \( 3r(3r+3) = 9r(r+1) \).
2. Sum \( = 9 \cdot \frac{n(n+1)(n+2)}{3} = 3n(n+1)(n+2) \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (c) \( n(n+1)(n+2) \)

Explanation:
1. The r-th term is \( 3r(r+1) \), and \( \sum r(r+1) = \frac{n(n+1)(n+2)}{3} \).
2. Sum = 3 x n(n+1)(n+2)/3 = n(n+1)(n+2).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (b) \( \dfrac{n(n+1)(3n^2 + 7n + 8)}{12} \)

Explanation:
1. Group the terms in threes: \( r^3 + r^2 + r \) for r = 1, ..., n.
2. Sum \( = \left[\frac{n(n+1)}{2}\right]^2 + \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(3n^2 + 7n + 8)}{12} \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

Answer: (a) \( \dfrac{n-3}{2n-5} \)

Explanation:
1. The sum to m terms is \( \frac{m}{2m+1} \).
2. With m = n - 3, the sum is \( \frac{n-3}{2(n-3)+1} = \frac{n-3}{2n-5} \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.

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

Explanation:
1. The r-th term is \( 2^r - 1 \).
2. Sum \( = (2 + 4 + \dots + 2^n) - n = 2^{n+1} - 2 - n \).

Teacher's Note:
1. Check the formula for n = 1 and n = 2 first, then confirm by the induction step: assume it for n = k and show it for n = k + 1.
2. Re-check the answer by substituting back into the question.