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- 64
- 36
- 49
- 25

**Answer: Option C**

: C

Substitute n=1 in23n−7n−1, we get

23−7−1=0→divisible by all positive integers Substitute n=2 in23n−7n−1, we get

26−14−1=49

LetP(n):23n−7n−1is divisible by 49

P(2) is true.

Assume P(k) is true

23k−7k−1=49m

Substituting k+1 in place of n, we get

23k+3−7(k+1)−1=8.23k−7k−8=8.(23k−7k−1)+7.7k=49(8m+k)→divisible by 49

P(k+1) is true

Hence, P(n) is true.

**Which of the following illustrates the inductive step to prove a statement P(n) about natural numbers n by mathematical induction, where k is an arbitrary natural number?**

__Question 2.__- P(1) is true
- P(k) is true
- P(k) is true ⇒ P(k+1) is true
- none of these

**Answer: Option C**

: C

Suppose there is a given statement P(n) involvingnatural numbers n such that i) The statement is true for a specific natural number m i.e. P(m) is true. This is known as the base case. ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n=k+1, i.e., the truth of P(k) implies the truth of P(k+1).

The second step is called the inductive step.

- True
- False
- 17
- 24

**Answer: Option A**

: A

P(n):4.7n+20isdivisibleby24

P(1):28+20=48→divisibleby24P(k):4.7k+20isdivisibleby24→assumetrue⇒4.7k+20=24qNow,4.7k+1+20=7.(4.7k)+20=6.4.7k+(4.7k+20)=24.7k+24qP(k+1)istrue

∴P(n)istrue4.7n+20isdivisibleby24

- 13
- 11
- 9
- none of these

**Answer: Option B**

: B

Substituting n=1, we have

101+1=11LetP(n):102n−1+1is divisible by 11

P(1) is true.

Assume P(k) is true

⇒102k−1+1=11mNow,P(k+1):102k+1+1=100.102k−1+1=99.102k−1+102k−1+111.(9.102k−1+m)→divisible by 11

P(k+1) is also true.

Hence, P(n) is true.

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