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Principle of Mathematical Induction

  • Last Updated : 31 Oct, 2020

Mathematical Induction is a mathematical proof method that is used to prove a given statement about any well-organized set. Generally, it is used for proving results or establishing statements that are formulated in terms of n, where n is a natural number. The technique involves three steps to prove a statement, P(n), as stated below:

  • Verify if the statement is true for trivial cases like n = a i.e. P(a) is true. [Base Case]
  • Assume that the statement is true for n = k for some k ≥ a i.e. P(k) is true. [Inductive Hypothesis]
  • If the truth of P(k) implies the truth of P(k + 1), then the statement P(n) is true for all n ≥ a.

The base step and the inductive step, together, prove that  P(k) => P(k + 1) => P(k + 2) …. is true. Therefore, P(n) is true for all integers n ≥ a. We can compare mathematical induction to falling dominoes. When a domino falls, it knocks down the next domino in succession. The first domino knocks down the second one, the second one knocks down the third, and so on. In the end, all of the dominoes will be bowled over. But there are some conditions to be fulfilled :

  1. The starting domino must fall to set the knocking process in action. This is the base step.
  2. The distance between dominoes must be equal for any two adjacent dominoes. Otherwise, a certain domino may fall down without bowling over the next. Then the sequence of reactions will stop. Maintaining the equal inter-domino distance ensures that  P(k) ⇒ P(k + 1) for each integer k ≥ a. This is the inductive step.

Examples

Example 1:  For all n ≥ 1, prove that,
12 + 22 + 32….n2 = {n(n + 1) (2n + 1)} / 6

Solution: 

Let the given statement be P(n),



P(n)=1^2+ 2^2 + 3^2+ ……+ n^2 = \frac{n(n + 1) (2n + 1)}{6}  \ for\, n=1,\ P(1)=\frac{1(1+1)(2*1+1)}{6} = 1\ ,which\ is\ true.

Now, let’s take a positive integer, k, and assume P(k) to be true i.e.,

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

We shall now prove that P(k + 1) is also true, so now we have,

P(k + 1) = P(k) + (k + 1)2  

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \\ \ \\ = \frac {k(k+1)(2k+1)+6{(k+1)}^2}{6} \\ \ \\ =  (k+1) \frac{( 2k^2 + k) + 6(k+1)}{6} \\ \ \\ =\frac{(k+1)(2k^2 +7k+6)}{6} \\ \ \\ =\frac{(k+1) (k+2) (2k+3)}{6}\\ \ \\ =\frac{(k+1) ((k+1)+1) (2(k+1) +1)}{6} \ \\

Thus P(k + 1) is true, whenever P(k) is true for all natural numbers. Hence, by the process of mathematical induction, the given result is true for all natural numbers.

Example 2: For all n ≥ 1, prove that,
1.2.3 + 2.3.4 + 3.4.5 …n(n + 1) (n + 2) = {n (n + 1) (n + 2) ( n + 3)} / 4



Solution: 

Let the given statement be S(n),

and \ S(n)=1.2.3+ 2.3.4 + 3.4.5+ ……+ n.(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n+3)}{4}\\ \ \\ for\ n=1, \ S(1) = \frac{1(1+1)(1+2)(1+3)}{4} = 6\ ,which\ is\ true. \ \\

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k)=1.2.3+ 2.3.4 + 3.4.5+ ……+ k.(k+1)(k+2) = \frac{k(k+ 1)(k + 2)(k+3)}{4} \ \\

We shall now prove that  S(k + 1) is also true, so now we have,

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

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 3: For all n ≥ 1, prove that,
1 + 3 + 5 … 2n – 1 = n2

Solution: 



Let the given statement be S(n), 

and S(n) = 1 + 3 + 5 … 2n – 1 = n2

For n = 1, 2 * 1 – 1 = 1 Thus S(1) is true .

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k) = 1+ 3 + 5 .. (2k – 1) = k

We shall now prove that  S(k + 1) is also true, so now we have,

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

L.H.S: 1 + 3 + 5 + …. (2k – 1 ) + 2k + 2 – 1

= S(k) + 2k + 1

= k2 + 2k + 1

= (k + 1)

= R.H.S

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 4: For all n ≥ 1, prove that,
1.2 + 2.3 + 3.4 …n(n + 1) = {n(n + 1)(n + 2)} / 3

Solution: 

Let the given statement be S(n),

and \ S(n)=1.2+ 2.3 + 3.4+ ……+ n.(n+1) = \frac{n(n + 1)(n + 2)}{3}\\ \ \\ for\ n=1, \ S(1) = \frac{1(1+1)(1+2)}{3} = 2\ ,which\ is\ true. \ \\

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k)=1.2+ 2.3 + 3.4+ ……+ k.(k+1) = \frac{k(k+ 1)(k + 2)}{3} \ \\

We shall now prove that S(k + 1) is also true, so now we have,



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

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 5:  Prove an = a1 + (n – 1) d, is the general term of any arithmetic sequence.

Solution: 

For n = 1, we have an = a1 + (1 – 1) d = a1, so the formula is true for n = 1,

Let us assume that the formula ak = a1 + (k – 1) is true for all natural numbers. 

We shall now prove that the formula is also true for k+1, so now we have,

ak + 1 = a1 + [(k + 1) – 1] d = a1 + k · d.

We assumed that ak = a1 + (k – 1) d, and by the definition of an arithmetic sequence ak+ 1 – ak = d,

then, ak + 1 – ak 

= (a1 + k · d) – (a1 + (k – 1)d)

= a1 – a1 + kd – kd + d

= d

Thus the formula is true for k + 1, whenever it is true for k. And we initially showed that the formula is true for n = 1. Thus the formula is true for all natural numbers.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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