RD Chapter 12 Mathematical Induction Ex 12.2 Solutions
Question - 21 : - 52n + 2 – 24n – 25is divisible by 576 for all n ϵ N
Answer - 21 : -
Let P (n): 52n + 2 – 24n – 25 is divisibleby 576
Let us check for n = 1,
P (1): 52.1+2 – 24.1 – 25
: 625 – 49
: 576
P (n) is true for n = 1. Where, P (n) is divisible by 576
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 52k + 2 – 24k – 25 is divisible by576
: 52k + 2 – 24k – 25 = 576λ …. (i)
We have to prove,
52k + 4 – 24(k + 1) – 25 is divisible by 576
5(2k + 2) + 2 – 24(k + 1) – 25 = 576μ
So,
= 5(2k + 2) + 2 – 24(k + 1) – 25
= 5(2k + 2).52 – 24k – 24– 25
= (576λ + 24k + 25)25 – 24k– 49 by using equation (i)
= 25. 576λ + 576k + 576
= 576(25λ + k + 1)
= 576μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 22 : - 32n + 2 – 8n – 9is divisible by 8 for all n ϵ N
Answer - 22 : -
Let P (n): 32n + 2 – 8n – 9 is divisible by8
Let us check for n = 1,
P (1): 32.1 + 2 – 8.1 – 9
: 81 – 17
: 64
P (n) is true for n = 1. Where, P (n) is divisible by 8
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 32k + 2 – 8k – 9 is divisible by 8
: 32k + 2 – 8k – 9 = 8λ … (i)
We have to prove,
32k + 4 – 8(k + 1) – 9 is divisible by 8
3(2k + 2) + 2 – 8(k + 1) – 9 = 8μ
So,
= 32(k + 1).32 – 8(k + 1) –9
= (8λ + 8k + 9)9 – 8k – 8 – 9
= 72λ + 72k + 81 – 8k – 17 using equation (1)
= 72λ + 64k + 64
= 8(9λ + 8k + 8)
= 8μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 23 : - (ab) n = an bn forall n ϵ N
Answer - 23 : -
Let P (n): (ab) n = an bn
Let us check for n = 1,
P (1): (ab) 1 = a1 b1
: ab = ab
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): (ab) k = ak bk …(i)
We have to prove,
(ab) k + 1 = ak + 1.bk+ 1
So,
= (ab) k + 1
= (ab) k (ab)
= (ak bk) (ab) using equation (1)
= (ak + 1) (bk + 1)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 24 : - n (n + 1) (n + 5) is a multiple of 3 for al ln ϵ N.
Answer - 24 : -
Let P (n): n (n + 1) (n + 5) is a multiple of 3
Let us check for n = 1,
P (1): 1 (1 + 1) (1 + 5)
: 2 × 6
: 12
P (n) is true for n = 1. Where, P (n) is a multiple of 3
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): k (k + 1) (k + 5) is a multiple of 3
: k(k + 1) (k + 5) = 3λ … (i)
We have to prove,
(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3
(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ
So,
= (k + 1) [(k + 1) + 1] [(k + 1) + 5]
= (k + 1) (k + 2) [(k + 1) + 5]
= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]
= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)
= 3λ + k2 + k + 2(k2 + 6k +5) + 2k + 2
= 3λ + k2 + k + 2k2 + 12k +10 + 2k + 2
= 3λ + 3k2 + 15k + 12
= 3(λ + k2 + 5k + 4)
= 3μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 25 : - 72n + 23n – 3.3n – 1 is divisible by 25 for all n ϵ N
Answer - 25 : -
Let P (n): 72n + 23n – 3. 3n – 1is divisible by 25
Let us check for n = 1,
P (1): 72 + 20.30
: 49 + 1
: 50
P (n) is true for n = 1. Where, P (n) is divisible by 25
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 72k + 23k – 3. 3k – 1 isdivisible by 25
: 72k + 23k – 3. 3k – 1 =25λ … (i)
We have to prove that:
72k + 1 + 23k. 3k isdivisible by 25
72k + 2 + 23k. 3k =25μ
So,
= 72(k + 1) + 23k. 3k
= 72k.71 + 23k. 3k
= (25λ – 23k – 3. 3k – 1) 49 + 23k.3k by using equation (i)
= 25λ. 49 – 23k/8. 3k/3. 49 + 23k.3k
= 24×25×49λ – 23k . 3k .49 + 24 . 23k.3k
= 24×25×49λ – 25 . 23k. 3k
= 25(24 . 49λ – 23k. 3k)
= 25μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.