Principles of Mathematical Induction: Important Questions

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Very Short Answer [1 Marks]

Ques. Prove that 1² + 2² + 3² + ---+ n² = \(\frac{n(n+1)(2n+1)}{6}\)

Ans.  Let P(n): 1² + 2² + 3² + ---+ n² = \(\frac{n(n+1)(2n+1)}{6}\)

For n= 1

L.H.S = 1² = 1

R.H.S = \(\frac{1(1+1) (2 \times 1+1)}{6} = \frac{1 \times 2 \times 3}{6} = 1\)

Since, L.H.S = R.H.S

∴ P(n) is true for n = 1

Proving P(k+1)  is true if P(k) is true

Assuming that P(k) is true, 

P (k) = 1 + 2² + 3² + … … + k² = \(\frac{k(k+1)(2k+1)}{6}\)   ...(1)

We will prove that P(k+1) is true. 

P (k+1) = 1 + 2² + 3² + … … + (k+1)² = \(\frac{(k+1)(k+1+1)(2 \times (k+1) +1)}{6}\)

P (k+1) = 1 + 2² + 3² + … … + (k+1)² = \(\frac{(k+1)(k+2)(2k + 2 + 1)}{6}\)

P (k+1) = 1 + 2² + 3² + … … + k² + (k+1)² = \(\frac{(k+1)(k+2)(2k + 3)}{6}\)

We have to prove P(k+1) is true.

Solving L.H.S, 

1 + 2² + 3² + … … + k² + (k+1)²

From (1): 1 + 2² + 3² + … … + k² = \(\frac{k(k+1)(2k+1)}{6}\)

= \(\frac{k(k+1)(2k+1)}{6} + (k+1)^2\)

= \(\frac{k(k+1)(2k+1) + 6 (k+1)^2}{6} \)

= \(\frac{(k+1) + (k(2k +1) + 6(k+1))}{6}\)

= \(\frac{(k+1)(2k^2 + k + 6k + 6)}{6}\)

= \(\frac{(k+1)(2k^2 + 7k + 6)}{6}\)

= \(\frac{(k+1)(2k^2 + 4k + 3k + 6)}{6}\)

= \(\frac{(k+1) + (k(2k +2) + 3(k+2))}{6}\)

= \(\frac{(k+1)(2k+3)(k+2)}{6}\)

= RHS

∴ Proving P(k+1)  is true if P(k) is true.

Thus, P(K+1) is true, whenever P(K) is true. Hence, from principles of mathematical induction, the statement P(n) is true for all natural numbers,n. 

Ques. Prove that 1 + 3 + 3² + -- + 3^n-1  = 3ⁿ  – 1 / 2 

Ans. Let P(n): 1 + 3 + 3² + -- + 3^n-1

= 3^{n – 1/2}

For n=1

L.H.S = 1

R.H.S = \(\frac{(3^1 - 1)}{2} = \frac{(3-1)}{2} = \frac{(2)}{2} = 1\)

Since, L.H.S = R.H.S

∴ P(n) is true for n = 1

Proving P(k+1)  is true if P(k) is true

Assuming that P(k) is true, 

P (k) = 1 + 3 + 3² + … … + 3^k-1 = \(\frac{(3^k - 1)}{2}\) ...(1)

We will prove that P(k+1) is true.

P(k+1) = 1 + 3 + 3² + … … + 3^{(k+1) -1} = \(\frac{(3^{k+1} - 1)}{2}\)

P(k+1) = 1 + 3 + 3² + …… + 3^(k-1) + 3^(k) = \(\frac{(3^{k+1} - 1)}{2}\)

We have to prove P(k+1) is true

Solving LHS

1 + 3 + 3² + …… + 3^(k-1) + 3^(k) 

From(1): 1 + 3 + 3² + …… + 3^{k – 1} = \(\frac{(3^k - 1)}{2}\)

= \(\frac{(3^k - 1)}{2}\) + 3^k

= \(\frac{(3^k - 1)+2 \times 3^k}{2}\)

= \(\frac{3^k + 2 \times 3^k - 1}{2}\)

= \(\frac{3(3^k) - 1}{2}\)

= \(\frac{3^{k+1} - 1}{2}\)

= RHS

∴ Proving P(k+1)  is true if P(k) is true.

Thus, By the principles of mathematical induction, the statement P(n) is true for n, where n is a natural number.

Ques. Prove by Principles of mathematical induction that (ab)ⁿ = aⁿ bⁿ

Ans. Let P(n) : (ab)ⁿ = aⁿ bⁿ

For n=1

ab=ab, which is true

Let P(K) be true

(ab)^k=a^kb^k….(1)

It is to be proved: P(K+1) is true

(ab)^k+1=a^k+1b^k+1

L.H.S = (ab)^K+1

= (ab)^k (ab)¹

= a^kb^k. (ab)¹ ……..from (1)

= a^k+1b^K+1

Thus, P(K+1) is true.

Ques. Prove that x²ⁿ – y²ⁿ is divisible by x + y.

Ans. Let P(n): x²ⁿ – y²ⁿ is divisible by x + y

For n=1

P(1)=x² – y²= (x-y) (x+y) is divisible by x + y

It is therefore true for, n=1

Let P(K) be true

P(K): x^2K - y^2K is divisible by x+y

= x^2K - y^2K = (x+y) λ, where λ= N (i)

We require to prove the result that is true for n=K+1

x2(K+1) - y2(K+1) = x^2K+2 - y^2K+2

= x^2K x² - y^2K y²

= ((x+y) λ + y^2K) x² - y^2K y² . . . (from i)

= (x+y) λ x² + x² y^2K - y^2K y²

= (x+y) λ x² + y^2K (x² - y²)

= (x+y) λ x² + y^2K (x+y) (x-y)

= (x+y) [x² λ+ y^2K (x-y)] is divisible by (x+y)

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, P(n) is true for all n????N.

Ques. Prove that n (n + 1) (2n + 1) is divisible by 6.

Ans. Let P(n): (n + 1) (2n + 1) is divisible by 6

For n=1

P(1)=(1) (2) (3)= 6 is divisible by 6.

Hence, the result is true for n=1

Let P(K) be true

 P(K): K(K+1) (2K+1) is divisible by 6

= K(K+1) (2K+1) = 6 λ, where λ= N (i)

We require to prove that the result obtained is true for n=K+1

(K+1) (K+2) (2K+3) = (K+1) (K+2) [(2K+1) +2]

= (K+1) (K+2) (2K+1) + 2(K+1) (K+2)

= (K+2) [(K+1) (2K+1)] + 2(K+1) (K+2)

= 6 λ + 2(K+1) (2K+1) + 2(K+1) (K+2) (by i)

= 6 λ + 2(K+1) [ 2K+1+K+2]

= 6 λ + 2(K+1) (3K+3)

= 6 λ + 6(K+1) (K+1)

= 6 [λ + (K+1) (K+1)] is divisible by 6

Hence, it can be said that (n + 1) (2n + 1) is divisible by 6.

Ques. Show that 2³ⁿ – 1 is divisible by 7.

Ans. Let P(n): 2³ⁿ – 1 is divisible by 7

For n=1

P(1)=2³ – 1= 7, which is divisible by 7

Hence, the result is true for n=1

Let P(K) be true

P(K): 2^3K-1 is divisible by 7

= 2^3K-1 = 7 λ, where λ= N (i)

We are required to prove that P(K+1) is true whenever P(K) is true.

2^3(K+1)-1= 2^3K+3 -1

=2^3K. 2³-1

=(7 λ+1).8-1 . . . (from i)

=56 λ+ 8 -1

= 56λ +7

= 7(8 λ+1) which is divisible by 7.

Thus, P(K+1) is true whenever P(K) is true. Hence, P(n) is true for all n∈N.

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Short Answers [2 Marks]

Ques. For every integer n, prove that 4 will be divisible to 7ⁿ – 3ⁿ.

Ans. P(n) : 7ⁿ­­­­-3ⁿ is divisible by 4

For n=1

P(1) : 7¹ -3¹ =4 which is divisible by 4

Thus, P(1) is true

Let P(k) be true

7^K – 3^K is divisible by 4λ, where λ=N…(i)

7^K – 3^K =4

We want to prove that P(K+1) is true, whenever P(K) is true.

7^K+1 – 3^K+1 = 7^K.7 – 3^K.3

=(4 λ +3^K). 7-3^K.3 . . . from (i)

= 28 λ +7.3^K - 3^K.3

= 28 λ + 3^K (7-3)

=4 (7 λ + 3^K)

Therefore, this proves that 7^K+1 – 3^K+1 is divisible by 4

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, the statement is true for every positive integer n. 

Ques. Prove the sum is multiple by 3: n(n+1) (n+5).

Ans. P(n) : n(n+1) (n+5) is multiple of 3

For n=1

P(1) : 1(1+1) (1+5) is multiple of 3

Thus, P(1) is true

Let P(k) be true

P(K)= K(K+1)(K+5) is multiple of 3

K(K+1) (K+5) =3λ, where λ=N….(i)

We are required to prove that the result is true for n=k+1

P(K+1) : (K+1) (K+2) (K+6)

= (K+1) (K+2) (K+6) = [ (K+1) (K+2)] (K+6)

= K(K+1) (K+2) + 6(K+1) (K+2)

=K(K+1) (K+5-3) + 6(K+1) (K+2)

=K(K+1) (K+5) – 3K(K+1) + 6(K+1) (K+2)

=K(K+1) (K+5) + (K+1) [6(K+2) – 3K]

=K(K+1) (K+5) + 3(K+1) (K+4)

=3 λ +3 (K+1) (K+4) ………from i

=3 [λ + (K+1) (K+4)] is a multiple of three

Hence, P(K+1) is a multiple of 3.

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, the statement is true for every positive integer n. 

Ques. Prove that 10^2n-1+1 is divisible by 11. 

Ans. P(n): 10^2n-1 +1 is divisible by 11, for n=1

P (1): 10^2-1 +1=11 is divisible by 11

Hence, the result is true for n=1

Let P(K) be true

P(K): 10^2n-1 +1 is divisible by 11

=10^2n-1 +1=11λ, where λ=N ……(i)

We are required to prove that the result is true for n=K+1

= 10^2(K+1)-1 +1=10^2K+2-1 +1

=10^2K+1 +1

=10^2K. 10¹ +1

= (110 λ-10).10+1 (from i)

=1100 λ-100+1

=1100 λ-99

=11(100 λ-9) is divisible by 11

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, P(n) is true for all n∈N.

Ques. Prove that: (2n+7) < (n+3)².

Ans. P(n) :(2n+7)<(n+3)²

For n=1

9 < (4)²

9<16, which is true

Let P(K) be true

(2K+7) < (K+3)²

Now, 2(K+1) + 7= (2K+7)+2 < (K+3)² +2 = K²+6K+11 < K² +8K+ 16= (K+4)2 = (K+3+1)²

Therefore, P(K+1) : 2(K+1) +7 < (K+1+3)²

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, P(n) is true for all n ???? N.

Ques. Prove that 41ⁿ – 14ⁿ can be multiplied by 27.

Ans. P(n) :41ⁿ – 14ⁿ is a multiple of 27

For n=1

P(1) : 41¹ – 14¹ =27 is a multiple of 27

Thus, P(1) is true

Let P(k) be true

P(K)= 41^K – 14^K is a multiple of 27

=41^K – 14^K =27λ, where λ=N…….(i)

We are required to prove that the result is true for n=K+1

41^K+1 – 14^K+1 = 41^K.41 -14^K.14

= (27λ +14^K).41-14^K.14……. From (i)

=27λ.41+14^K.41-14^K..14

=27λ.41+14^K (41-14)

= 27λ.41+14^K (27)

=27(41λ+14^K) is a multiple of 27

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, P(n) is true for all n∈N.

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Long Answers [3 Marks]

Ques. Prove that the sum of cubes of n natural numbers is equal to (n(n+1)²)² for all n natural numbers.

Ans. In the given statement we are asked to prove:

1³+2³+3³+ . . . +n³ = (n(n+1)²)²

Step 1: Now with the help of the principle of induction in math let us check the validity of the given statement P(n) for n=1.

P(1)=(1(1+1)²)² = 1 This is true.

Step 2: Now as the given statement is true for n=1 we shall move forward and try proving this for n=k, i.e.,

1³+2³+3³+….+k³= (k(k+1)²)².

Step 3: Let us now try to establish that P(k+1) is also true.

1³+2³+3³+...+k³+(k+1)³ = (k(k+1)²)²+(k+1)³

⇒1³+2³+3³+...+k³+(k+1)³ =k²(k+1)4+(k+1)³

= k²(k+1)² + 4((k+1)³)⁴

=(k+1)²(k²+4(k+1))⁴

=(k+1)²(k²+4k+4)⁴

= (k+1)²((k+2)²)⁴

=(k+1)²(k+1+1)²)⁴

=(k+1)²((k+1)+1)²)⁴

Ques. Show that 1 + 3 + 5 + … + (2n−1) = n²

Ans.  Step 1: Result is true for n = 1

That is 1 = (1)² (True)

Step 2: Assume that result is true for n = k

1 + 3 + 5 + … + (2k−1) = k²

Step 3: Check for n = k + 1

i.e. 1 + 3 + 5 + … + (2(k+1)−1) = (k+1)²

We can write the above equation as,

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

Using step 2 result, we get

k² + (2(k+1)−1) = (k+1)²

k² + 2k + 2 −1 = (k+1)²

k² + 2k + 1 = (k+1)²

(k+1)² = (k+1)²

L.H.S. and R.H.S. are the same.

So the result is true for n = k+1

The statement is true, in accordance with Mathematical induction.

It is simultaneously true for n=k+1. Hence, this statement is valid for all natural numbers n.

Ques. Demonstrate that 2²ⁿ-1 is divisible by 3 via principles of mathematical induction.

Ans. To prove: 2²ⁿ-1 is divisible by 3

Assume that the given statement be P(k)

TIt can otherwise be claimed as P(k) = 2²ⁿ-1 is divisible by 3, for every natural number

Step 1: In step 1, assume n= 1, so that the given statement can be written as

P(1) = 2²⁽¹⁾-1 = 4-1 = 3. So 3 is divisible by 3. (i.e.3/3 = 1)

Step 2: Now, assume that P(n) is true for all the natural number, say k

Hence, the given statement can be written as

P(k) = 2^2k-1 is divisible by 3.

It means that 22k-1 = 3a (where a belongs to natural number)

Now, we need to prove the statement is true for n= k+1

Hence, P(k+1) = 2^2(k+1)-1

P(k+1)= 2^2k+2-1

P(k+1)= 2^2k. 22 – 1

P(k+1)= (2^2k. 4)-1

P(k+1) =3.2k +(2^2k-1)

The above expression can be written as

P(k+1) =3.2^2k + 3a

Now, take 3 outside, we get

P(k+1) = 3(2^2k + a)= 3b, where “b” belongs to natural number

It is proved that p(k+1) holds true, whenever the statement P(k) is true.

Thus, 2²ⁿ-1 is divisible by 3 is proved using the principles of mathematical induction

Ques. Prove that 2n > n for every positive integer ‘n’.

Ans. Assume that P(n): 2n > n

If n =1, 2¹>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

2^k > k …(1)

Elsewise, P(k +1) is true whenever P(k) is true.

Now, multiplying both sides of the equation (1) by 2, we get

2. 2^k > 2k

Now by using the property,

i.e., 2^k+1> 2k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Ques. Prove that 1 + 3 + 5 + … + (2n – 1) = n² via principle of Mathematical induction.

Ans. Given Statement: 1 + 3 + 5 + … + (2n – 1) = n²

Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n² , for n ∈ N

Note that P(1) is true, since

P(1) : 1 = 1²

Let P(k) is true for some k ∈ N,

It means that,

P(k) : 1 + 3 + 5 + … + (2k – 1) = k²

To prove that P(k + 1) is true, we have

1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k² + (2k + 1)

= k² + 2k + 1

By using the formula, the above form can be written as:

= (k + 1)²

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

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Very Long Answers [5 Marks]

Ques. Prove that 12ⁿ + 25^n-1 is divisible by 13.

Ans. P(n) : 12n + 25^n-1 is divisible by 13

For n=1

P(1): 12 + (25)0 = 13

Let P(K) be true

P(K): 12^K + 25^K-1 is divisible by 13

12^K +25^K-1 = 13λ . . . (i)

We are required to prove that the result is true for n=K+1

12K+1 + 25^K = 12^K.121+25^K

= (13λ -25^K-1).12+25^K . . . from (i)

= 13X 12 λ +25^K-1 .12+25^K

=13 X 12 λ +25^K-1 (-12+25)

=13 (12 λ+25^K-1) which is divisible by 13.

Thus, P(K+1) is true whenever P(K) is true. Hence, by principles of mathematical induction, P(n) is true for all n∈N.

Ques. Prove 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133.

Ans. P(n) = 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133.

For n=1

P(1): 11³ +12³ =3059

Which is divisible by 133

Let P(K) be true

P(K) : 11^K+2 +12^2K+1 is divisible by 133

=11^K+2 +12^2K+1 =133λ . . . (i)

We are required to prove that the result is true for n=K+1

L.H.S = 11^K+1+2 +12^2(K+1)+1

=11^K+3 +12^2K+2+1

=11^K+3 +12^2K+3

=11^K.11³ +12^2K. 12³

=11^K+2 .11 +12^2K.12³

=133 x 11λ -12^2K+1.11+12^2K.12³ . . . from (i)

=133 x 11λ -12^2K+1 .11+12^2K.12³

=133 x 11 x λ +12^K (-12 x 11 +12³)

=133 [(11 λ+12^K (1596)] which is divisible by 133

Thus, P(K+1) is true whenever P(K) is true. Hence, P(n) is proved to be true for all n∈N.

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