RD Chapter 12 Mathematical Induction Ex 12.2 Solutions
Question - 11 : - 2 + 5 + 8 + 11 + … + (3n – 1) =1/2 n (3n + 1)
Answer - 11 : -
Let P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)
Let us check for n = 1,
P (1): 2 = 1/2 × 1 × 4
: 2 = 2
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) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)
So,
2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)
Now, substituting the value of P (k) we get,
= 1/2 × k (3k + 1) + (3k + 2) by using equation (i)
= [3k2 + k + 2 (3k + 2)] / 2
= [3k2 + k + 6k + 2] / 2
= [3k2 + 7k + 2] / 2
= [3k2 + 4k + 3k + 2] / 2
= [3k (k+1) + 4(k+1)] / 2
= [(k+1) (3k+4)] /2
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 12 : - 1.3 + 2.4 + 3.5 + … + n. (n+2) =1/6 n (n+1) (2n+7)
Answer - 12 : -
Let P (n): 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1)(2n+7)
Let us check for n = 1,
P (1): 1.3 = 1/6 × 1 × 2 × 9
: 3 = 3
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): 1.3 + 2.4 + 3.5 + … + k. (k+2) = 1/6 k (k+1) (2k+7) …(i)
So,
1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)
Now, substituting the value of P (k) we get,
= 1/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)
= (k+1) [{k(2k+7)/6} + {(k+3)/1}]
= (k+1) [(2k2 + 7k + 6k + 18)] / 6
= (k+1) [2k2 + 13k + 18] / 6
= (k+1) [2k2 + 9k + 4k + 18] / 6
= (k+1) [2k(k+2) + 9(k+2)] / 6
= (k+1) [(2k+9) (k+2)] / 6
= 1/6 (k+1) (k+2) (2k+9)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 13 : - 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n+ 1) = n(4n2 + 6n – 1)/3
Answer - 13 : -
Let P (n): 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) =n(4n2 + 6n – 1)/3
Let us check for n = 1,
P (1): (2.1 – 1) (2.1 + 1) = 1(4.12 + 6.1-1)/3
: 1×3 = 1(4+6-1)/3
: 3 = 3
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): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = k(4k2 +6k – 1)/3 … (i)
So,
1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)
Now, substituting the value of P (k) we get,
= k(4k2 + 6k – 1)/3 + (2k + 1) (2k + 3) byusing equation (i)
= [k(4k2 + 6k-1) + 3 (4k2 +6k + 2k + 3)] / 3
= [4k3 + 6k2 – k + 12k2 +18k + 6k + 9] /3
= [4k3 + 18k2 + 23k + 9] /3
= [4k3 + 4k2 + 14k2 +14k +9k + 9] /3
= [(k+1) (4k2 + 8k +4 + 6k + 6 – 1)] / 3
= [(k+1) 4[(k+1)2 + 6(k+1) -1]] /3
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 14 : - 1.2 + 2.3 + 3.4 + … + n(n+1) = [n(n+1) (n+2)] / 3
Answer - 14 : -
Let P (n): 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1)(n+2)] / 3
Let us check for n = 1,
P (1): 1(1+1) = [1(1+1) (1+2)] /3
: 2 = 2
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): 1.2 + 2.3 + 3.4 + … + k(k+1) = [k (k+1) (k+2)] / 3 …(i)
So,
1.2 + 2.3 + 3.4 + … + k(k+1) + (k+1) (k+2)
Now, substituting the value of P (k) we get,
= [k (k+1) (k+2)] / 3 + (k+1) (k+2) by using equation (i)
= (k+2) (k+1) [k/2 + 1]
= [(k+1) (k+2) (k+3)] /3
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 15 : - 1/2 + 1/4 + 1/8 + … + 1/2n =1 – 1/2n
Answer - 15 : -
Let P (n): 1/2 + 1/4 + 1/8 + … + 1/2n =1 – 1/2n
Let us check for n = 1,
P (1): 1/21 = 1 – 1/21
: 1/2 = 1/2
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.
Let P (k): 1/2 + 1/4 + 1/8 + … + 1/2k =1 – 1/2k … (i)
So,
1/2 + 1/4 + 1/8 + … + 1/2k + 1/2k+1
Now, substituting the value of P (k) we get,
= 1 – 1/2k + 1/2k+1 by usingequation (i)
= 1 – ((2-1)/2k+1)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 16 : - 12 + 32 +52 + … + (2n – 1)2 = 1/3 n (4n2 –1)
Answer - 16 : -
Let P (n): 12 + 32 + 52 +… + (2n – 1)2 = 1/3 n (4n2 – 1)
Let us check for n = 1,
P (1): (2.1 – 1)2 = 1/3 × 1 × (4 – 1)
: 1 = 1
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): 12 + 32 + 52 +… + (2k – 1)2 = 1/3 k (4k2 – 1) … (i)
So,
12 + 32 + 52 +… + (2k – 1)2 + (2k + 1)2
Now, substituting the value of P (k) we get,
= 1/3 k (4k2 – 1) + (2k + 1)2 byusing equation (i)
= 1/3 k (2k + 1) (2k – 1) + (2k + 1)2
= (2k + 1) [{k(2k-1)/3} + (2k+1)]
= (2k + 1) [2k2 – k + 3(2k+1)] / 3
= (2k + 1) [2k2 – k + 6k + 3] / 3
= [(2k+1) 2k2 + 5k + 3] /3
= [(2k+1) (2k(k+1)) + 3 (k+1)] /3
= [(2k+1) (2k+3) (k+1)] /3
= (k+1)/3 [4k2 + 6k + 2k + 3]
= (k+1)/3 [4k2 + 8k – 1]
= (k+1)/3 [4(k+1)2 – 1]
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 17 : - a + ar + ar2 + … +arn – 1 = a [(rn – 1)/(r – 1)], r ≠ 1
Answer - 17 : -
Let P (n): a + ar + ar2 + … + arn– 1 = a [(rn – 1)/(r – 1)]
Let us check for n = 1,
P (1): a = a (r1 – 1)/(r-1)
: a = a
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): a + ar + ar2 + … + ark – 1 =a [(rk – 1)/(r – 1)] … (i)
So,
a + ar + ar2 + … + ark – 1 +ark
Now, substituting the value of P (k) we get,
= a [(rk – 1)/(r – 1)] + ark byusing equation (i)
= a[rk – 1 + rk(r-1)] / (r-1)
= a[rk – 1 + rk+1 – r‑k]/ (r-1)
= a[rk+1 – 1] / (r-1)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 18 : - a + (a + d) + (a + 2d) + … + (a +(n-1)d) = n/2 [2a + (n-1)d]
Answer - 18 : -
Let P (n): a + (a + d) + (a + 2d) + … + (a + (n-1)d) =n/2 [2a + (n-1)d]
Let us check for n = 1,
P (1): a = ½ [2a + (1-1)d]
: a = a
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): a + (a + d) + (a + 2d) + … + (a + (k-1)d) = k/2 [2a +(k-1)d] … (i)
So,
a + (a + d) + (a + 2d) + … + (a + (k-1)d) + (a + (k)d)
Now, substituting the value of P (k) we get,
= k/2 [2a + (k-1)d] + (a + kd) by using equation (i)
= [2ka + k(k-1)d + 2(a+kd)] / 2
= [2ka + k2d – kd + 2a + 2kd] / 2
= [2ka + 2a + k2d + kd] / 2
= [2a(k+1) + d(k2 + k)] / 2
= (k+1)/2 [2a + kd]
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 19 : - 52n – 1 isdivisible by 24 for all n ϵ N
Answer - 19 : -
Let P (n): 52n – 1 is divisible by 24
Let us check for n = 1,
P (1): 52 – 1 = 25 – 1 = 24
P (n) is true for n = 1. Where, P (n) is divisible by 24
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 52k – 1 is divisible by 24
: 52k – 1 = 24λ … (i)
We have to prove,
52k + 1 – 1 is divisible by 24
52(k + 1) – 1 = 24μ
So,
= 52(k + 1) – 1
= 52k.52 – 1
= 25.52k – 1
= 25.(24λ + 1) – 1 by using equation (1)
= 25.24λ + 24
= 24λ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
Question - 20 : - 32n + 7 isdivisible by 8 for all n ϵ N
Answer - 20 : -
Let P (n): 32n + 7 is divisible by 8
Let us check for n = 1,
P (1): 32 + 7 = 9 + 7 = 16
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 + 7 is divisible by 8
: 32k + 7 = 8λ
: 32k = 8λ – 7 … (i)
We have to prove,
32(k + 1) + 7 is divisible by 8
32k + 2 + 7 = 8μ
So,
= 32(k + 1) + 7
= 32k.32 + 7
= 9.32k + 7
= 9.(8λ – 7) + 7 by using equation (i)
= 72λ – 63 + 7
= 72λ – 56
= 8(9λ – 7)
= 8μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.