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RD Chapter 12- Mathematical Induction Ex-12.2 Interview Questions Answers

Question 1 :
Prove the following by the principle of mathematical induction:
1. 1 + 2 + 3 + … + n = n (n +1)/2 i.e., the sum of the first n natural numbers is n (n + 1)/2.

Answer 1 :

Let us consider P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2

For, n = 1

LHS of P (n) = 1

RHS of P (n) = 1 (1+1)/2 = 1

So, LHS = RHS

Since, P (n) is true for n = 1

Let us consider P (n) be the true for n = k, so

1 + 2 + 3 + …. + k = k (k+1)/2 … (i)

Now,

(1 + 2 + 3 + … + k) + (k + 1) = k (k+1)/2 + (k+1)

= (k + 1) (k/2 + 1)

= [(k + 1) (k + 2)] / 2

= [(k+1) [(k+1) + 1]] / 2

P (n) is true for n = k + 1

P (n) is true for all n N

So, by the principle of Mathematical Induction

Hence, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2 is true forall n N.

Question 2 :

12 + 22 +32 + … + n2 = [n (n+1) (2n+1)]/6

Answer 2 :

Let us consider P (n) = 12 + 22 +32 + … + n2 = [n (n+1) (2n+1)]/6

For, n = 1

P (1) = [1 (1+1) (2+1)]/6

1 = 1

P (n) is true for n = 1

Let P (n) is true for n = k, so

P (k): 12 + 22 + 32 +… + k2 = [k (k+1) (2k+1)]/6

Let’s check for P (n) = k + 1, so

P (k) = 12 + 22 + 32 +– – – – – + k2 + (k + 1)2 = [k + 1 (k+2)(2k+3)] /6

= 12 + 22 + 32 +– – – – – + k2 + (k + 1)2

= [k + 1 (k+2) (2k+3)] /6 + (k + 1)2

= (k +1) [(2k2 + k)/6 + (k + 1)/1]

= (k +1) [2k2 + k + 6k + 6]/6

= (k +1) [2k2 + 7k + 6]/6

= (k +1) [2k2 + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 3 :

1 + 3 + 32 + … + 3n-1 =(3n – 1)/2

Answer 3 :

Let P (n) = 1 + 3 + 32 + – – – – + 3n –1 = (3n – 1)/2

Now, For n = 1

P (1) = 1 = (31 – 1)/2 = 2/2 =1

P (n) is true for n = 1

Now, let’s check for P (n) is true for n = k

P (k) = 1 + 3 + 32 + – – – – + 3k – 1 =(3k – 1)/2 … (i)

Now, we have to show P (n) is true for n = k + 1

P (k + 1) = 1 + 3 + 32 + – – – – + 3k =(3k+1 – 1)/2

Then, {1 + 3 + 32 + – – – – + 3k – 1}+ 3k + 1 – 1

= (3k – 1)/2 + 3k using equation (i)

= (3k – 1 + 2×3k)/2

= (3×3 k – 1)/2

= (3k+1 – 1)/2

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 4 :

1/1.2 + 1/2.3 + 1/3.4 + … +1/n(n+1) = n/(n+1)

Answer 4 :

Let P (n) = 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

For, n = 1

P (n) = 1/1.2 = 1/1+1

1/2 = 1/2

P (n) is true for n = 1

Let’s check for P (n) is true for n = k,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2) =(k+1)/(k+2)

Then,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2)

= 1/(k+1)/(k+2) + k/(k+1)

= 1/(k+1) [k(k+2)+1]/(k+2)

= 1/(k+1) [k2 + 2k + 1]/(k+2)

=1/(k+1) [(k+1) (k+1)]/(k+2)

= (k+1) / (k+2)

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 5 :

1 + 3 + 5 + … + (2n – 1) = n2 i.e.,the sum of first n odd natural numbers is n2.

Answer 5 :

Let P (n): 1 + 3 + 5 + … + (2n – 1) = n2

Let us check P (n) is true for n = 1

P (1) = 1 =12

1 = 1

P (n) is true for n = 1

Now, Let’s check P (n) is true for n = k

P (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)

We have to show that

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2

Now,

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k2 + (2k + 1)

= k2 + 2k + 1

= (k + 1)2

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 6 :

1/2.5 + 1/5.8 + 1/8.11 + … +1/(3n-1) (3n+2) = n/(6n+4)

Answer 6 :

Let P (n) = 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) =n/(6n+4)

Let us check P (n) is true for n = 1

P (1): 1/2.5 = 1/6.1+4 => 1/10 = 1/10

P (1) is true.

Now,

Let us check for P (k) is true, and have to prove that P (k+ 1) is true.

P (k): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1) (3k+2) =k/(6k+4)

P (k +1): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1)(3k+2) +1/(3k+3-1)(3k+3+2)

: k/(6k+4) + 1/(3k+2)(3k+5)

: [k(3k+5)+2] / [2(3k+2)(3k+5)]

: (k+1) / (6(k+1)+4)

P (k + 1) is true.

Hence proved by mathematical induction.

Question 7 :

1/1.4 + 1/4.7 + 1/7.10 + … +1/(3n-2)(3n+1) = n/3n+1

Answer 7 :

Let P (n) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) =n/3n+1

Let us check for n = 1,

P (1): 1/1.4 = 1/4

1/4 = 1/4

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) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k-2)(3k+1) = k/3k+1… (i)

So,

[1/1.4 + 1/4.7 + 1/7.10 + … +1/(3k-2)(3k+1)]+ 1/(3k+1)(3k+4)

= k/(3k+1) + 1/(3k+1)(3k+4)

= 1/(3k+1) [k/1 + 1/(3k+4)]

= 1/(3k+1) [k(3k+4)+1]/(3k+4)

= 1/(3k+1) [3k2 + 4k + 1]/ (3k+4)

= 1/(3k+1) [3k2 + 3k+k+1]/(3k+4)

= [3k(k+1) + (k+1)] / [(3k+4) (3k+1)]

= [(3k+1)(k+1)] / [(3k+4) (3k+1)]

= (k+1) / (3k+4)

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 8 :

1/3.5 + 1/5.7 + 1/7.9 + … +1/(2n+1)(2n+3) = n/3(2n+3)

Answer 8 :

Let P (n) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) =n/3(2n+3)

Let us check for n = 1,

P (1): 1/3.5 = 1/3(2.1+3)

: 1/15 = 1/15

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) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) =k/3(2k+3) … (i)

So,

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) +1/[2(k+1)+1][2(k+1)+3]

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/(2k+3)(2k+5)

Now substituting the value of P (k) we get,

= k/3(2k+3) + 1/(2k+3)(2k+5)

= [k(2k+5)+3] / [3(2k+3)(2k+5)]

= (k+1) / [3(2(k+1)+3)]

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

Question 9 :

1/3.7 + 1/7.11 + 1/11.15 + … +1/(4n-1)(4n+3) = n/3(4n+3)

Answer 9 :

Let P (n) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) =n/3(4n+3)

Let us check for n = 1,

P (1): 1/3.7 = 1/(4.1-1)(4+3)

: 1/21 = 1/21

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): 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) =k/3(4k+3) …. (i)

So,

1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) +1/(4k+3)(4k+7)

Substituting the value of P (k) we get,

= k/(4k+3) + 1/(4k+3)(4k+7)

= 1/(4k+3) [k(4k+7)+3] / [3(4k+7)]

= 1/(4k+3) [4k2 + 7k +3]/ [3(4k+7)]

= 1/(4k+3) [4k2 + 3k+4k+3] / [3(4k+7)]

= 1/(4k+3) [4k(k+1)+3(k+1)]/ [3(4k+7)]

= 1/(4k+3) [(4k+3)(k+1)] / [3(4k+7)]

= (k+1) / [3(4k+7)]

P (n) is true for n = k + 1

Hence, P (n) is true for all n  N.

Question 10 :

1.2 + 2.22 + 3.23 +… + n.2= (n–1) 2n + 1 + 2

Answer 10 :

Let P (n) = 1.2 + 2.22 + 3.23 +… + n.2= (n–1) 2n + 1 + 2

Let us check for n = 1,

P (1):1.2 = 0.20 + 2

: 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 + 2.22 + 3.23 +… + k.2= (k–1) 2k + 1 + 2 …. (i)

So,

{1.2 + 2.22 + 3.23 + … + k.2k}+ (k + 1)2k + 1

Now, substituting the value of P (k) we get,

= [(k – 1)2k + 1 + 2] + (k + 1)2k + 1 usingequation (i)

= (k – 1)2k + 1 + 2 + (k + 1)2k + 1

= 2k + 1(k – 1 + k + 1) + 2

= 2k + 1 × 2k + 2

= k × 2k + 2 + 2

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.


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