RD Chapter 12 Mathematical Induction Ex 12.1 Solutions
Question - 1 : - If P (n) is the statement тАЬn (n + 1) is evenтАЭ, then what is P (3)?
Answer - 1 : -
Given:
P (n) = n (n + 1) is even.
So,
P (3) = 3 (3 + 1)
= 3 (4)
= 12
Hence, P (3) = 12, P (3) is also even.
Question - 2 : - If P (n) is the statement тАЬn3┬а+n is divisible by 3тАЭ, prove that P (3) is true but P (4) is not true.
Answer - 2 : -
Given:
P (n) = n3┬а+ n is divisible by 3
We have┬аP (n) = n3┬а+ n
So,
P (3) = 33┬а+ 3
= 27 + 3
= 30
P (3) = 30, So it is divisible by 3
Now, letтАЩs check with P (4)
P (4) = 43┬а+ 4
= 64 + 4
= 68
P (4) = 68, so it is not divisible by 3
Hence, P (3) is true and P (4) is not true.
Question - 3 : - If P (n) is the statement тАЬ2n┬атЙе3nтАЭ, and if P (r) is true, prove that P (r + 1) is true.
Answer - 3 : -
Given:
P (n) = тАЬ2n┬атЙе 3nтАЭ and p(r) is true.
We have,┬аP (n) = 2n┬атЙе 3n
Since, P (r) is true
So,
2rтЙе┬а3r
Now, letтАЩs multiply both sides by 2
2├Ч2rтЙе┬а3r├Ч2
2r + 1тЙе┬а6r
2r + 1тЙе┬а3r + 3r [since 3r>3 = 3r + 3rтЙе3 +3r]
тИ┤ 2r + 1тЙе┬а3(r + 1)
Hence, P (r + 1) is true.
Question - 4 : - If P (n) is the statement тАЬn2┬а+ nтАЭis evenтАЭ, and if P (r) is true, then P (r + 1) is true
Answer - 4 : -
Given:
P (n) = n2┬а+ n is even and P (r) is true,then r2┬а+ r is even
Let us consider r2┬а+ r = 2k тАж (i)
Now, (r + 1)2┬а+ (r + 1)
r2┬а+ 1 + 2r + r + 1
(r2┬а+ r) + 2r + 2
2k + 2r + 2 [from equation (i)]
2(k + r + 1)
2╬╝
тИ┤┬а(r + 1)2┬а+ (r + 1) is Even.
Hence, P (r + 1) is true.
Question - 5 : - Given an example of a statement P (n) such that it is true for all n ╧╡ N.
Answer - 5 : -
Let us consider
P (n) = 1 + 2 + 3 + тАУ тАУ тАУ тАУ тАУ + n = n(n+1)/2
So,
P (n) is true for all natural numbers.
Hence, P (n) is true for all n тИИ N.
Question - 6 : - If P (n) is the statement тАЬn2┬атАУ n +41 is primeтАЭ, prove that P (1), P (2) and P (3) are true. Prove also that P(41) is not true.
Answer - 6 : -
Given:
P(n) = n2┬атАУ n + 41 is prime.
P(n) = n2┬атАУ n + 41
P (1) = 1 тАУ 1 + 41
= 41
P (1) is Prime.
Similarly,
P(2) = 22┬атАУ 2 + 41
= 4 тАУ 2 + 41
= 43
P (2) is prime.
Similarly,
P (3) = 32┬атАУ 3 + 41
= 9 тАУ 3 + 41
= 47
P (3) is prime
Now,
P (41) = (41)2┬атАУ 41 + 41
= 1681
P (41) is not prime
Hence, P (1), P(2), P (3) are true but P (41) is not true.