MATHEMTICAL INDUCTION

 

Introduction to Mathematical Induction

Mathematical induction is a fundamental proof technique in mathematics, often used to establish the veracity of an infinite number of cases. It is particularly useful for proving statements about integers, sequences, and series. The method relies on proving a base case and then showing that if the statement holds for an arbitrary case, it must hold for the next one. This creates a domino effect, proving the statement for all subsequent cases.

The Principle of Mathematical Induction

The principle of mathematical induction can be summarized in two main steps:

1. Base Case: Verify that the statement is true for the initial value, usually $n = 0$ or $n = 1$. This step provides the foundation for the induction.

   For example, if we are proving a statement $P(n)$ for all $n \geq 1$, we first show that $P(1)$ is true.

2. Inductive Step: Assume the statement is true for some arbitrary integer $k$, i.e., assume $P(k)$ is true. Then, prove that $P(k+1)$ is also true under this assumption. This step shows that if the statement holds for one case, it holds for the next.

   Using our example, we assume $P(k)$ is true and then prove $P(k+1)$.

If both steps are successfully completed, by the principle of mathematical induction, the statement $P(n)$ is true for all integers $n \geq 1$.

MATHEMTICAL INDUCTION

QUESTIONS

1.    Using the principle of mathematical induction, prove that 

1² + 2² + 3² + .... + n² = (1/6) {n (n + 1) (2n + 1} for all n ∈ N. 

Solution:

 Here, for n=1

P (1) = 1² + 2² + 3² + .... + 1² = (1/6) {1(1+1) (2(2) +1)}

                                           = 1

                                  L.H.S = R.H.S

Hence P(n) is true for n = 1

Let P(k) is true.

P(k): 1² + 2² + 3² + .... + k² = (1/6) {k(k+1) (2k+1)} …….1    

Proving P(k+1) is true.

P(k+1): 1² + 2² + 3² + .... + k²+ (k+1)² = (1/6) {(k+1)(k+2)(2k+3)}

From (1):  1² + 2² + 3² + .... + k² = (1/6) {k(k+1) (2k+1)}

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

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

                                                 = (1/6) {(k+1)(2k²+k+6k+6)}

                                                             = (1/6) {(k+1)(2k²+7k+6)}

                                                 = (1/6) {(k+1)(2k²+4k+3k+6)}

                                                 = (1/6) {(k+1)(2k(k+2)+3(k+2))}

                                                 = (1/6) {(k+1)(2k+3)(k+2)}                                                                             

 

 

2.    Using the principle of mathematical induction, prove that

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}

Solution:

Here, for n=1

L.H.S:        = 1.2 = 2

R.H.S:        = (1/3) {n(n + 1)(n + 2)}= (1.2.3)/3 = 6/3 = 2

                             L.H.S = R.H.S

For n = k we will show that :

          1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k + 1) = (1/3){k(k+1)(k+2)}…..1

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

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + (k+1).((k+1)+1) = (1/3){(k+1)((k+1)+1)((k+1)+2}

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + (k+1)(k+2) = (1/3){(k+1)(k+2)(k+3}

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k+1)+(k+1)(k+2) = (1/3){(k+1)(k+2)(k+3)}

We have to prove P(k+1) from P(k)

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k.(k+1) = (1/3){k(k+1)(k+2)}

Adding (k+1).(k+2) on both sides we, will have

1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k+1)+(k+1)(k+2) =(1/3){k(k+1)(k+2)}+(k+1)(k+2)

                                                = (1/3){k(k+1)(k+2)+3(k+1)(k+2)

                                                = (1/3) (k+1)(k+2)(k+3)

Thus 1.2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k+1)+(k+1)(k+2) = (1/3){(k+1)(k+2)(k+3)}

Which is the same as P(k+1)

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

 

 

 

3. By using mathematical induction prove that the given equation is true for all positive integers.

2 + 4 + 6 + …. + 2n = n(n+1)

Solution:

for n = 1

Here ,P (1) 2+4+6+….+2(1) = 1(1+1)

                                      2=2

                             L.H.S= R.H.S

For n= k

                  

            2+4+6+….+2(k) = k(k+1)

          2+4+6+….+2k   = k(k+1)…..1

 

For n= k+1

          2+4+6+….+2(k+1) = (k+1){(k+1)+1}

          2+4+6+….+2(k+1) = (k+1)(k+2)

Now Adding for P(k+1) and P(k)

          2+4+6+….+2(k+1)+2k = k(k+1)+2(k+1)

 

                                               =(k+1) (k+2)

                                      L.H.S = R.H.S

Hence, P(n) is true for n = (k+1)

 

4.    Using the principle of mathematical induction, prove that

1.3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2n - 1)(2n + 1) = (1/3){n(4n² + 6n - 1).

Solution:

For n = 1

                        1.3=3

                        = (1(4(1)²+(6(1)-1) / 3

                        = ((4+5)) / 3

                        = 9 / 3

                        L.H.S=R.H.S

For n = k

   1.3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2k- 1)(2k + 1) = (1/3){k(4k² + 6k - 1)

 

For n = k+1 , we will have

On L.H.S, we will have ,

   1.3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2k+1)(2k + 3)

Now adding the values of k and k+1 we will have:

   1.3 + 3 ∙ 5 + 5 ∙ 7 +.....(2k-1)(2k+1)+ (2k+1)(2k + 3)

Now by the induction Hypothesis on the R.H.S we will have :

Adding on the R.H.S

 

= (1/3){k(4k² + 6k - 1)} + (2k+1)+(2k+3)

= (1/3){k(4k² + 6k - 1) + 3 (2k+1)+(2k+3)

= (1/3) {4k³+6k²-k+12k²+24k+9}

= (1/3) {4k³+18k²+23k+9}

= (1/3) {4k³+6k²-k+12k²+24k+9}

= (1/3) {4(k+1)²+6(k+1)-1}

This is the optimal solution for the particular values of k & k+1

 

5. By using mathematical induction prove that the given equation is true for all positive integers.

2 + 6 + 10 + ….. + (4n - 2) = 2n² 

Solution:

For n = 1

                   2 = 2(1)²

                   2 = 2

                 L.H.S = R.H.S

For n = k

2 + 6 + 10 + ….. + (4k - 2) = 2k²

For n = k+1

2 + 6 + 10 + ….. + (4k + 2) = 2(k+1)²

Adding the values of k and k+1

2 + 6 + 10 + ….. +(4k+2)+ (4k - 2) = 2k²+(4(k+1)-2)

                                                      = 2k²+ (4k+4 -2)

                                                      =2k² + 4k+2

                                                      =2(k² + 2k+1)

                                                      =2(k+1)²

                                      L.H.S = R.H.S

Hence the values of k and k+1 are same for each other.