The document discusses the principle of mathematical induction and how it can be used to prove statements about natural numbers. It provides examples of using induction to prove statements about sums, products, and divisibility. The principle of induction states that to prove a statement P(n) is true for all natural numbers n, one must show that P(1) is true and that if P(k) is true, then P(k+1) is also true. The document provides examples of direct proofs of P(1) and inductive proofs of P(k+1) to demonstrate applications of the principle.

3. 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.

4.  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.

5.  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.

6.  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.

7. •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.

8. 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.

9. 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]

10. 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.

11. Q2. Prove the following by using the principle of
mathematical induction for all n ∊ N
1.2.3+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)/4
For n=1
P(1)≣ 1.2.3= 1(1+1)(1+2)(1+3)/4
6 = 1×2×3×4/4
6 = 24/4=6 which is true.
therefore, P(n) is true, where n=1
Assume that P(k) is true for some positive integer k

12. =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 true
Now 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)/4
Thus P(k+1) is true, whenever P(k) is true
Hence, from the principle of mathematical
induction, the statement P(n) is true for all
natural numbers n.

13. 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.

14. 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.

15. 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.

16. Now we wish to prove that P(k+1) is true
whenever 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 3
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.

17. 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]

18. 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.

19. 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]

20. Consider the identity
=(1+x)ᵏ⁺¹
=(1+x)ᵏ(1+x)
Given that x> -1, so (1+x) > 0
Therefore, 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 of
mathematical induction the statement P(n)
is true for all natural numbers.