Natural numbers

Natural Numbers: Introduction, Mathematical Induction, Variants of Induction, Induction with Non-zero Base Cases.

 

Ques.Describe mathematical induction.

Answer:

Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.

The technique involves two steps to prove a statement, as stated below −

Step 1(Base step) − It proves that a statement is true for the initial value.

Step 2(Inductive step) − It proves that if the statement is true for the nth iteration (or number n), then it is also true for (n+1)th iteration ( or number n+1).

Method:

Step 1: Consider an initial value for which the statement is true. It is to be shown that the statement is true for n = initial value.

Step 2: Assume the statement is true for any value of n = k. Then prove the statement is true for n = k+1. We actually break n = k+1 into two parts, one part is n = k (which is already proved) and try to prove the other par

 

Ques.Prove that n2 + 2n is divisible by 3 using principle of mathematical induction, where n is natural number.

AKTU 2015-16, Marks 10

Answer:

 

Ques.Prove by induction: 

AKTU 2016-17, Marks 10

Answer:

 

Ques. Prove by the principle of mathematical induction, that the sum of finite number of terms of a geometric progression, a + ar + ar2 + …… arn-1 = a(rn - 1)/(r-1) if r ≠ 1.

AKTU 2014-15, Marks 05

Answer:

 

Ques. Prove that if n is a positive integer, then 133 divides 11n+1 + 122n-1.

AKTU 2018-19, Marks 07

Answer:

Ques. Prove by mathematical induction n4 - 4n2 is divisible by 3 for all n > = 2.

AKTU 2017-18, Marks 07

Answer:

 Assume the inductive hypothesis is true for n. 

We need to show that (n+1)4 −4(n+1)2 is divisible by 3.

 By the inductive hypothesis, we know that n 4 −4n 2 is divisible by 3.

 Hence (n+1)4 −4(n+1)2 is divisible by 3 if (n+1)4 −4(n+1)2 −(n 4 −4n 2 ) is divisible by 3. Now 

(n + 1)4 − 4(n + 1)2 − (n 4 − 4n 2 ) 

= n 4 + 4n 3 + 6n 2 + 4n + 1 − 4n 2 − 8n − 4 − n 4 + 4n2 

= 4n 3 + 6n 2 − 4n − 3, 

which is divisible by 3 if 4n 3−4n is.

 Since 4n 3−4n = 4n(n+1)(n−1), we see that 4n3−4n is always divisible by 3.

 Going backwards, we conclude that (n+1)4 −4(n+1)2 is divisible by 3, and that the inductive hypothesis holds for n + 1.

 By the Principle of Mathematical Induction, n 4 − 4n 2 is divisible by 3, for all n ∈ N

 

Ques.  Prove by mathematical induction 

3 + 33 + 333 + ………. 3333 = (10n+1 - 9n - 10)/27

AKTU 2017-18, Marks 07

Answer:

Let P(n) be the given statement

Now,

P(n): 3+33+333+...+333..n digits..3=(10n+1−9n−10)/27

Step I: 

P(1)=3=(102−9−10)/27=1/27×81

Thus, P(1) is true.

Step II:Let P(m) be true.

Then,

3+33+333+...+333..m digits..3=(10m+1−9m−10)/27

We need to show that P(m+1) is true whenever P(m) is true.

Now, P(m+1)=3+33+333+...+333..(m+1)digits...3

This is a geometric progression with n=m+1.

∴Sum P(m+1)

=[9+99+999+...(m+1)term]/3

=[(10−1)+(100−1)+..(m+1)term]/3

=[10+100+1000+...(m+1)term − (1+1+1...(m+1)times...+1)]/3

=[10(10m+1−1)/9−m+1]/3=[10m+2−9m−19]/27

Thus, P(m+1) is true.

By the principle of mathematical induction, P(n) is true for all n∈N.