principle of mathematical induction

In algebra or in other discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To prove such statements well-suited principle that is used-based on the specific technique is know as the principle of mathematical induction.

The principle of mathematical induction is

one such tool which can be used to prove a

wide variety of mathematical statements.

Each such statement is assumed as P(n)

associated with positive integer n, for which

the correctness for the case n=1 is examined.

Then assuming the truth of P(k) for some

positive integer k, the truth of P(k+1) is

established.

There is a given statement P(n) involving the natural number n such that

(i) The statement is true for n=1, i.e., P(1) is

true, and

(ii) If the statement is true for n=k (where k is

some positive integer ), then the statement

is also true for n=k+1 , i.e., truth of P(k)

implies the truth of P(k+1).

(iii) Then, P(n) is true for all natural numbers

n.

One key basis for mathematical thinking is deductive reasoning.

Deduction in a nutshell is given a statement to be proven, often called a conjecture or a theorem in mathematics, valid deductive steps are derived and a proof may or may not be established, i.e., deduction is the application of a general case to a particular case.

Deduction: Generalization → Specific Instances

One such example is:

(a) Eight is divisible by two.

(b) Any number which is divisible by two is an even number. Therefore,

(c) Eight is an even number.

•To understand the basic principle of mathematical induction,s suppose a set of thin rectangular tiles are placed in order as shown in figure.•When the first tile is pushed in the indicated direction •(a) The first tile falls, and•(b) In the event that any tile falls its successor will necessarily falls.•This is underlying principle of mathematical induction.

Statements giving expression about summation or multiplication of special series.

Statements to show the divisibility of an expression by a certain natural number.

Statements containing signs of inequality.

Q1.For all n ≥ 1, prove that

1²+2²+3²+4²+……..+n²= n(n+1)(2n+1)/6

Ans:Let the given statement be P(n)

P(n):1²+2²+3²+4²+……+n²= n(n+1)(2n+1)/6

For n=1,

P(1)≣1=1(1+1)(2×1+1)/6=1×2×3/6=1 which is

true

therefore, P(n) is true. Where n = 1

Assume that P(k) is true for some positive integer

k ,i.e.,

1²+2²+3²+4²+……..+k²=k(k+1)(2k+1)/6 --------[1]

We shall now prove that P(k+1) is also true.

Now we have,

(1²+2²+3²+4²+……..+k²)+(k+1)²

= k(k+1)(2k+1)/6+(k+1)² {Using[1]}

= k(k+1)(2k+1)+6(k+1)²/6

= (k+1)(2k+7k+6)/6

= (k+1)(k+1+1){2(k+1)+1}/6

Thus P(k+1) is true, wherever P(k) is true.

Hence, from the principle of mathematical induction , the statement is true for all natural number n.

Q2. Prove the following by using the principle of mathematical induction for all n ∊ N1.2.3+2.3.4+……..+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

Ans : Let the given statement be P(n)P(n)=1.2.3+2.3.4+………..+n(n+1)(n+2)=

n(n+1)(n+2)(n+3)/4For n=1P(1)≣ 1.2.3= 1(1+1)(1+2)(1+3)/4

6 = 1×2×3×4/46 = 24/4=6 which is true.

therefore, P(n) is true, where n=1Assume that P(k) is true for some positive integer k

=1.2.3+2.3.4+………+k(k+1)(k+2)=k(k+1)(k+2)(k+3)/4 -------------------[1]

We shall now prove that P(k+1) is also trueNow we have,=1.2.3+2.3.4+………..+(k+1)(k+1+1)(k+1+2)= k(k+1)(k+2)(k+3)/4 + (k+1)(k+2)(k+3)= k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)/4= (k+1)(k+2)(k+3)(k+4)/4Thus P(k+1) is true, whenever P(k) is trueHence, from the principle of mathematical induction, the statement P(n) is true for allnatural numbers n.

Q1. For every positive integer n, prove that

7ⁿ- 3ⁿ is divisible by 4

Ans: We can write

P(n): 7ⁿ+3ⁿ is divisible by 4

We note that

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

Thus P(n) is true for n=1

Let P(k) be true for some natural number k

i.e., P(k) : 7ᵏ-3ᵏ is divisible by 4.

We can write 7ᵏ-3ᵏ = 4d, where d ∊ N.

Now we wish to prove that P(k+1) is true

whenever P(k) is true.

Now,

=7⁽ᵏ⁺¹⁾-3⁽ᵏ⁺¹⁾=7⁽ᵏ⁺¹⁾-7.3ᵏ+7.3ᵏ-3⁽ᵏ⁺¹⁾

=7(7ᵏ-3ᵏ)+(7-3)3ᵏ

=7(4d)+(7-3)3ᵏ

=7(4d) +4.3ᵏ

=4(7d+3ᵏ)

From the last line, we see that 7⁽ᵏ⁺¹⁾-3⁽ᵏ⁺¹⁾ is

divisible by 4. Thus P(k+1) is true when P(k)

is true. Therefore, by principle of

mathematical induction the statement P(n) is

true for every positive integer n.

Q2.Prove the following by using the principle

of mathematical induction for al n ∊ N

n(n+1)(n+5) is a multiple of 3

Ans: We can write

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

We note that

P(1): 1(1+1)(1+5) = 12 which is a multiple of3

Thus P(n) is true for n=1

Let P(k) be true for some natural number k

i.e., P(k) : k(k+1)(k+5) is a multiple of 3

We can write, k(k+1)(k+5) = 3d, where d ∊ N.

Now we wish to prove that P(k+1) is truewhenever P(k) is true.Now,=(k+1)(k+1+1)(k+1+5)=(k+1)[k(k+5) + k×1 +2(k+5) + 2×1]=(k+1)[k(k+5) + k+ 2k + 10+ 2]=(k+1)[k(k+5) +3k +12]=k(k+1)(k+5)+(k+1)3(k+4)=3d + 3(k+1)(k+4)=3[d + (k+1)(k+4)]From the last line, we see that (k+1)(k+1+1)(k+1+5) is a multiple of 3Thus P(k+1) is true when P(k)is true. Therefore, by principle ofmathematical induction the statement P(n) is true for every positive integer n.

Q1. Prove that 2ⁿ>n for all positive integers n.

Ans: Let P(n) be the given statement

P(n): 2ⁿ>n

when n=1,

2¹>n.

Hence, P(1) is true

Assume that P(k) is true for any positive

integer k, i.e.,

2ᵏ>k -------------------[1]

We shall now prove that P(k+1) is true

whenever P(k) is true.

Multiplying both sides of [1] by 2, we get

=2.2ᵏ>2k

i.e.,

=2ᵏ⁺¹>2k

=k+k>k+1

Therefore P(k+1) is true when P(k) is true.

Hence, by principle of mathematical

induction, P(n) is true for every positive

integer n.

Q2. Prove that (1+x)ⁿ≥ (1+nx), for all natural number n, where x> -1.

Ans: Let P(n) be the given statement,

P(n): (1+x)ⁿ≥(1+nx), for x> -1

We know that

(1+x)≥(1+x) for x> -1

P(n) is true when n=1

Assume that

P(k): (1+x)ᵏ≥(1+kx), x> -1 is true. ---------------[1]

We want to prove that P(k+1) is true for x> -1

whenever P(k) is true. ---------------[2]

Consider the identity=(1+x)ᵏ⁺¹=(1+x)ᵏ(1+x)Given that x> -1, so (1+x) > 0Therefore, by using (1+x)ᵏ ≥ (1+kx), we have=(1+x)ᵏ⁺¹ ≥ (1+kx)(1+x)=(1+x)ᵏ⁺¹ ≥ (1+x+kx+kx²)Here k is a natural number and x²≥ 0 so that kx² ≥ 0. Therefore=(1+x+kx+kx²) ≥ (1+x+kx)and so we obtained=(1+x)ᵏ⁺¹ ≥ (1+x+kx)=(1+x)ᵏ⁺¹ ≥ [1+(1+k)x]Thus the statement in [2] is established .Hence by principle ofmathematical induction the statement P(n) is true for all natural numbers.