InterviewSolution

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1.	What will be P(k + 1) for P(n) = n^3 (n + 1)?(a) (k + 1)^4(b) k^4 + 5k^3 + 9k^2 + 7k + 2(c) k^4 + 6k^3 + 9k^2 + 7k + 2(d) k^4 + 3k^3 + 9k^2 + 6k + 2I have been asked this question in my homework.I would like to ask this question from The Principle of Mathematical Induction in section Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct CHOICE is (b) k^4 + 5k^3 + 9k^2 + 7k + 2 The BEST I can EXPLAIN: P(N) = n^3 (n + 1) P(k + 1) = (k + 1)^3 (k + 1 + 1) P(k + 1) = (k^3 + 3k^2 + 3k + 1) (k + 2) P(k + 1) = k^4 + 3k^3 + 3k^2 + k + 2k^3 + 6k^2 + 6k + 2 P(k + 1) = k^4 + 5k^3 + 9k^2 + 7k + 2

2.	For principle of mathematical induction to be true, what type of number should ‘n’ be?(a) Whole number(b) Natural number(c) Rational number(d) Any form of numberThe question was posed to me by my college professor while I was bunking the class.Query is from The Principle of Mathematical Induction in chapter Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct choice is (a) WHOLE number To explain: According to the PRINCIPLE of Mathematical induction, X(N) can be true if X(1) is true and if X(K) is true. When X(k) is true, it implies that X(k + 1) is also true. Here n can be equal to 0, 1, 2, 3 and so on.

3.	n^3 + 5n is divisible by which of the following?(a) 3(b) 5(c) 7(d) 11The question was asked in unit test.This is a very interesting question from The Principle of Mathematical Induction topic in chapter Principle of Mathematical Induction of Mathematics – Class 11
Answer» Right answer is (a) 3 To elaborate: P(N) = n^3 + 5n P(1) = 1 + 5 P(1) = 6 We assume the P(k) is true and divisible by 6. P(k) = k^3 + 5k is divisible by 6 and can be written as 6c or 3 x 2C We need to PROVE that P(k + 1) is divisible by 6 P(k + 1) = (k + 1)^3 + 5(k + 1) P(k + 1) = k^3 + 1 + 3k^2 + 3k + 5k + 5 P(k + 1) = (k^3 + 5k) + 3k^2 + 3k + 6 P(k + 1) = 6c + 3(k^2 + k + 2) P(k + 1) = (3 x 2c) + 3(k^2 + k + 2) Therefore, P(k + 1) is definitely divisible by 3

4.	State whether the following series is true or not.(a) }{2}\)(b) True(c) FalseThis question was posed to me in exam.My doubt is from The Principle of Mathematical Induction topic in division Principle of Mathematical Induction of Mathematics – Class 11
Answer» Right OPTION is (a) }{2}\) For EXPLANATION: P(N) = n(n + 1)/2 P(1) = 1 We assume P(k) to be true, therefore, P(k) = \(\frac{k(k + 1)}{2}\) To prove that, P(k + 1) = \(\frac{(k+1)(k+ 2)}{2}\) PROOF: P(k + 1) = 1 + 2 + 3 +….+ k + k + 1 P(k + 1) = \(\frac{k(k + 1)}{2}\) + k+1 P(k + 1) = \(\frac{k(k+ 1)+2(k+1)}{2}\) P(k + 1) = \(\frac{(k+1)(k+ 2)}{2}\) Therefore, P(n) is true by principle of mathematical INDUCTION.

5.	What would be the hypothesis of mathematical induction for n(n + 1) < n! (where n ≥ 4) ?(a) It is assumed that at n = k, k(k + 1)! > k!(b) It is assumed that at n = k, k(k + 1)! < k!(c) It is assumed that at n = k, k(k + 1)! > (k + 1)!(d) It is assumed that at n = k, (k + 1)(k + 2)! < k!The question was posed to me in a national level competition.The query is from The Principle of Mathematical Induction topic in portion Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct option is (b) It is assumed that at n = k, k(k + 1)! < k! For explanation I would say: When we use the principle of MATHEMATICAL induction, we ASSUME that P(n) is TRUE for P(k) and prove that P(k + 1) is also true. Here P(k) is k(k + 1)! < k!

6.	If P(k) = k^2 (k + 3) (k^2 – 1) is true, then what is P(k + 1)?(a) (k + 1)^2 (k + 3) (k^2 – 1)(b) (k + 1)^2 (k + 4) (k^2 – 1)(c) (k + 1)^2 (k + 4) k (k + 2)(d) (k + 1) (k + 4) k (k +2)I had been asked this question in unit test.Asked question is from The Principle of Mathematical Induction in portion Principle of Mathematical Induction of Mathematics – Class 11
Answer» CORRECT choice is (c) (K + 1)^2 (k + 4) k (k + 2) Easiest explanation: In mathematical induction, if P(k) is true, we need to prove that P(k + 1) is also true. Here P(k + 1) is FOUND by substituting (k + 1) in place of k. P(k + 1) = (k + 1)^2 (k + 1 + 3) ((k + 1)^2 – 1) P(k + 1) = (k + 1)^2 (k + 4) (k^2 + 1 + 2k – 1) P(k + 1) = (k + 1)^2 (k + 4) (k^2 + 2k) P(k + 1) = (k + 1)^2 (k + 4) k (k +2)

7.	n^2 + 3n is always divisible by which number, provided n is an integer?(a) 2(b) 3(c) 4(d) 5I had been asked this question in my homework.The doubt is from The Principle of Mathematical Induction topic in chapter Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct OPTION is (a) 2 Easy EXPLANATION: P(n) = n^2 + 3n P(1) = 1 + 3 P(1) = 4 Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k^2 + 3k can be written as 4c. We need to check if P(k + 1) is divisible by 4 P(k+1) = (k + 1)^2 + 3(k + 1) P(k+1) = k^2 + 1 + 2k + 3k + 3 P(k+1) = k^2 + 5k + 4 P(k+1) = (k^2 + 3k) + 2k + 4 P(k+1) = 4c + 2k + 4 P(k+1) = 4c + 2(k + 2) Clearly the second part of the equation is not divisible by 4. However P(k) = 4c is divisible by 2 and P(k + 1) is ALSO divisible by 2. Therefore, 2 divides P(n).

8.	P(n) = n(n^2 – 1). Which of the following does not divide P(k+1)?(a) k(b) k + 2(c) k + 3(d) k + 1This question was posed to me in class test.My question is from The Principle of Mathematical Induction in division Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct answer is (C) K + 3 Easy explanation: P(N) = n(n^2 – 1) P(k + 1) = (k + 1) ((k + 1)^2 – 1) P(k + 1) = (k + 1) (k^2 + 1 + 2k – 1) P(k + 1) = (k + 1) (k^2 + 2k) P(k + 1) = (k + 1) k (k + 2) Therefore, k, (k + 1), (k – 1) divide P(k + 1).

9.	If 10^3n + 2^4k + 1. 9 + k, is divisible by 11, then what is the least positive value of k?(a) 7(b) 6(c) 8(d) 10I have been asked this question at a job interview.I would like to ask this question from The Principle of Mathematical Induction topic in portion Principle of Mathematical Induction of Mathematics – Class 11
Answer» Correct answer is (d) 10 To elaborate: P(n) = 10^3n + 2^4k + 1. 9 + k P(1) = 10^3 + 2^5 . 9 + k P(1) = 1000 + 288 + k P(1) = 1288 + k When 1288 is DIVIDED by 11, the REMAINDER is 1. Therefore, 1287 is divisible by 11. The next number that is divisible is 1298. k = 1298 – 1288 k = 10

10.	By principle of mathematical induction, 2^4n-1 is divisible by which of the following?(a) 8(b) 3(c) 5(d) 7I have been asked this question in unit test.Enquiry is from The Principle of Mathematical Induction in section Principle of Mathematical Induction of Mathematics – Class 11
Answer» The correct option is (a) 8 Best explanation: P(n) = 2^4n – 1 P(1) = 2^3 = 8 Let us assume P(k) is divisible by 8 and can be written as 8C, where C is any integer. P(k) = 2^4k – 1 = 8c P(k + 1) = 2^4(k + 1) – 1 P(k + 1) = 2^4k + 3 P(k + 1) = 2^4 . 2^4k – 1 P(k + 1) = 2^4 . 8c Clearly, P(k + 1) is divisible by 2, 4, 8 and 16.

11.	7^2n + 2^2n – 2 . 3^n – 1 is divisible by 50 by principle of mathematical induction.(a) True(b) FalseI got this question in an online interview.This question is from The Principle of Mathematical Induction in section Principle of Mathematical Induction of Mathematics – Class 11
Answer» Right CHOICE is (b) False Easy explanation: P(n) = 7^2n + 2^2n – 2 . 3^n – 1 P(1) = 7^2 + 2^0 . 3^0 P(1) = 50 We now ASSUME that P(k) is divisible by 50. Therefore, P(k) = 7^2k + 2^2k – 2 . 3^k – 1 To PROVE P(k + 1) = 7^2(k – 1) + 2^2(k + 1) – 2 . 3^(k + 1) – 1 is divisible by 50 P(k + 1) = 7^2k . 7^2 + 2^2k . 3^k P(k + 1) = 7^2 ( 7^2k + 2^2n – 2 . 3^k – 1 – 2^2k – 2. 3^k – 1 ) + 2^2k . 3^k P(k + 1) = 7^2 x 50c – 7^2 . 2^2k – 2 . 3^k – 1 + 2^2k . 3^k Since P(k) = 7^2k + 2^2n – 2 . 3^k – 1 – 2^2k – 2 . 3^k – 1 is divisible by 50, it can be written as 50c. P(k + 1) =49 x 50C – 49 . 2^2k – 2 . 3^k – 1 + 3 x 4 x 2^2k – 2 . 3^k – 1 P(k + 1) = 49 x 50C – 2^2k – 2 . 3^k – 1 x 37 While the first term is divisible by 50, the second term is not. Therefore, by principle of mathematical induction, 7^2n + 2^2n – 2 . 3^n – 1 is not divisible by 50.