Question 1: If P (n) is the statement “n (n + 1) is even”, then what is P (3)?
Solution:
Given: P (n) = n (n + 1) is even.
Substituting n with 3 we get,
P (3) = 3 (3 + 1)
P (3) = 3 (4)
P (3) = 12
Since P (3) = 12, and 12 is even
Therefore, P (3) is also even.
Question 2: If P (n) is the statement “n3 + n is divisible by 3”, prove that P (3) is true but P (4) is not true.
Solution:
Given: P (n) = n3 + n is divisible by 3
Firstly, substituting n with 3 we get,
P (3) = 33 + 3
P (3) = 27 + 3
P (3) = 30
Since P (3) = 30, and 30 is divisible by 3
Therefore, P (3) is true.
Now, substituting n with 4 we get,
P (4) = 43 + 4
P (4) = 64 + 4
P (4) = 68
Since P (4) = 68, and 68 it is not divisible by 3
Question 3: If P (n) is the statement “2n ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.
Solution:
Given: P (n) = 2n ≥ 3n and p(r) is true.
also given that P (r) is true
When we substitute n with r we get
P (r) = 2r ≥ 3r
Now, multiply both sides by 2 we get,
2 × 2r ≥ 3r × 2
2r + 1 ≥ 6r
We can write 6r as 3r + 3r
2r + 1 ≥ 3r + 3r
Since 3r ≥ 3, therefore 3r + 3r ≥ 3 + 3r
Substituting 3r + 3r with 3 + 3r we get,
2r + 1 ≥ 3 + 3r
2r + 1 ≥ 3(r + 1) [Taking 3 as common]
Since, 2r+1 ≥ 3(r + 1) is equal to P (r + 1)
Therefore, P (r + 1) is true.
Question 4: If P (n) is the statement “n2 + n is even”, and if P (r) is true, then P (r + 1) is true
Solution:
Given: P (n) = n2 + n is even
Also given that P (r) is true,
Therefore, P (r) = r2 + r is even
Let us consider r2 + r = 2x … (i)
Substituting r with r + 1
Now, (r + 1)2 + (r + 1)
r2 + 1 + 2r + r + 1 [ formula = (a + b)2 = a2 + 2ab + b2]
(r2 + r) + 2r + 2
2x + 2r + 2 [from equation (i) we get 2x = r2 + r]
2(x + r + 1)
2μ
Since, (r + 1)2 + (r + 1) is Even which is equal to P (r + 1).
Therefore, P (r + 1) is true.
Question 5: Given an example of a statement P (n) such that it is true for all n ϵ N.
Solution:
Let us consider P (n) as
P (n) = 1 + 2 + 3 + – – – – – + n = n(n+1)/2
Since P (n) is true for all natural numbers.
Therefore, P (n) is true for all n ∈ N.
Question 6: If P (n) is the statement “n2 – n + 41 is prime”, prove that P (1), P (2), and P (3) are true. Prove also that P (41) is not true.
Solution:
Given: P(n) = n2 – n + 41 is prime.
Substituting n with 1, we get
P (1) = 1 – 1 + 41
P (1) = 41
Since P (1) = 41, and 41 is prime.
Therefore, P (1) is true.
Now substituting n with 2, we get
P(2) = 22 – 2 + 41
P(2) = 4 – 2 + 41
P(2) = 43
Since P (2) = 43, and 43 is prime.
Therefore, P (2) is true.
Now substituting n with 3, we get
P (3) = 32 – 3 + 41
P (3) = 9 – 3 + 41
P (3) = 47
Since P (3) = 47, and 47 is prime.
Therefore, P (3) is true.
Now substituting n with 41, we get
P (41) = (41)2 – 41 + 41
P (41) = 1681
Since P (41) = 1681, and 1681 is not prime.
Therefore, P (41) is not true.
Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape,
GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out -
check it out now!