Mathematics Questions and Answers – The Principle of Mathematical Induction

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “The Principle of Mathematical Induction”.

1. 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 number
View Answer

Answer: a
Explanation: 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.
advertisement

2. 72n + 22n – 2 . 3n – 1 is divisible by 50 by principle of mathematical induction.
a) True
b) False
View Answer

Answer: b
Explanation: P(n) = 72n + 22n – 2 . 3n – 1
P(1) = 72 + 20 . 30
P(1) = 50
We now assume that P(k) is divisible by 50.
Therefore, P(k) = 72k + 22k – 2 . 3k – 1
To prove P(k + 1) = 72(k – 1) + 22(k + 1) – 2 . 3(k + 1) – 1 is divisible by 50
P(k + 1) = 72k . 72 + 22k . 3k
P(k + 1) = 72 ( 72k + 22n – 2 . 3k – 1 – 22k – 2 . 3k – 1 ) + 22k . 3k
P(k + 1) = 72 x 50c – 72 . 22k – 2 . 3k – 1 + 22k . 3k
Since P(k) = 72k + 22n – 2 . 3k – 1 – 22k – 2 . 3k – 1 is divisible by 50, it can be written as 50c.
P(k + 1) = 49 x 50C – 49 . 22k – 2 . 3k – 1 + 3 x 4 x 22k – 2 . 3k – 1
P(k + 1) = 49 x 50C – 22k – 2 . 3k – 1 x 37
While the first term is divisible by 50, the second term is not.
Therefore, by principle of mathematical induction, 72n + 22n – 2 . 3n – 1 is not divisible by 50.

3. By principle of mathematical induction, 24n-1 is divisible by which of the following?
a) 8
b) 3
c) 5
d) 7
View Answer

Answer: a
Explanation:
P(n) = 24n – 1
P(1) = 23 = 8
Let us assume P(k) is divisible by 8 and can be written as 8c, where c is any integer.
P(k) = 24k – 1 = 8c
P(k + 1) = 24(k + 1) – 1
P(k + 1) = 24k + 3
P(k + 1) = 24 . 24k – 1
P(k + 1) = 24 . 8c
Clearly, P(k + 1) is divisible by 2, 4, 8 and 16.
advertisement
advertisement

4. If 103n + 24k + 1. 9 + k, is divisible by 11, then what is the least positive value of k?
a) 7
b) 6
c) 8
d) 10
View Answer

Answer: d
Explanation: P(n) = 103n + 24k + 1. 9 + k
P(1) = 103 + 25 . 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

5. P(n) = n(n2 – 1). Which of the following does not divide P(k+1)?
a) k
b) k + 2
c) k + 3
d) k + 1
View Answer

Answer: c
Explanation: P(n) = n(n2 – 1)
P(k + 1) = (k + 1) ((k + 1)2 – 1)
P(k + 1) = (k + 1) (k2 + 1 + 2k – 1)
P(k + 1) = (k + 1) (k2 + 2k)
P(k + 1) = (k + 1) k (k + 2)
Therefore, k, (k + 1), (k – 1) divide P(k + 1).
advertisement

6. 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!
View Answer

Answer: b
Explanation: 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!

7. If P(k) = k2 (k + 3) (k2 – 1) is true, then what is P(k + 1)?
a) (k + 1)2 (k + 3) (k2 – 1)
b) (k + 1)2 (k + 4) (k2 – 1)
c) (k + 1)2 (k + 4) k (k + 2)
d) (k + 1) (k + 4) k (k +2)
View Answer

Answer: c
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) (k2 + 1 + 2k – 1)
P(k + 1) = (k + 1)2 (k + 4) (k2 + 2k)
P(k + 1) = (k + 1)2 (k + 4) k (k +2)
advertisement

8. n2 + 3n is always divisible by which number, provided n is an integer?
a) 2
b) 3
c) 4
d) 5
View Answer

Answer: a
Explanation: P(n) = n2 + 3n
P(1) = 1 + 3
P(1) = 4
Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k2 + 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) = k2 + 1 + 2k + 3k + 3
P(k+1) = k2 + 5k + 4
P(k+1) = (k2 + 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).

9. n3 + 5n is divisible by which of the following?
a) 3
b) 5
c) 7
d) 11
View Answer

Answer: a
Explanation: P(n) = n3 + 5n
P(1) = 1 + 5
P(1) = 6
We assume the P(k) is true and divisible by 6.
P(k) = k3 + 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) = k3 + 1 + 3k2 + 3k + 5k + 5
P(k + 1) = (k3 + 5k) + 3k2 + 3k + 6
P(k + 1) = 6c + 3(k2 + k + 2)
P(k + 1) = (3 x 2c) + 3(k2 + k + 2)
Therefore, P(k + 1) is definitely divisible by 3
advertisement

10. State whether the following series is true or not.
1 + 2 + 3 +…..+ n = \(\frac{n(n + 1)}{2}\)
a) True
b) False
View Answer

Answer: a
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.

11. What will be P(k + 1) for P(n) = n3 (n + 1)?
a) (k + 1)4
b) k4 + 5k3 + 9k2 + 7k + 2
c) k4 + 6k3 + 9k2 + 7k + 2
d) k4 + 3k3 + 9k2 + 6k + 2
View Answer

Answer: b
Explanation: P(n) = n3 (n + 1)
P(k + 1) = (k + 1)3 (k + 1 + 1)
P(k + 1) = (k3 + 3k2 + 3k + 1) (k + 2)
P(k + 1) = k4 + 3k3 + 3k2 + k + 2k3 + 6k2 + 6k + 2
P(k + 1) = k4 + 5k3 + 9k2 + 7k + 2

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter