Question

Prove by the principle of mathematical induction that for all n € N, n² + n is even natural number.


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Answer 1

In math induction proof we will work on some examples using mathematical induction.

Induction proof is a mathematical method of proving a set of formula or theory or series of natural numbers. Induction proof is used from the theory of mathematical induction which is similar to the incident of fall of dominoes. When we push a domino in a set of dominoes the falling of first domino leads to the falling of other dominoes. So if there is a set of infinite dominoes then a falling effect of single domino will lead to the falling of other dominoes one by one. Now to prove this incident we can say the first domino falls to second and the second will fall to third and this way others will fall. Similarly in induction proof for infinite series of n numbers set where P (n) is the set property, we do not need to prove the property for all natural numbers. We prove it for two instances. First is for P (0) or P (1) and this step is called base step. In the second step we need to assume first that the property P (k) is true which is called induction hypothesis. By using inductive hypothesis we need to prove the P (k+1) is true and this step is called inductive step. This is how in induction proof for property P (n), we need to prove it for P (0) or P (1) and P (k+1) and then it can be stated that the property statement is true or holds for all the natural numbers.

Now we will try to understand induction proof from an example. First we take a property of sum of n natural numbers.

1 + 2 + 3 + ……. + n = n(n+1)2

The above set of natural numbers is property P (n) which is simply a formula of sum of n natural numbers. By using induction proof technique we need to prove that this formula holds true for all natural numbers. As stated before the first step is base step P (1).

For P (1),  

LHS = 1

RHS = 1(1+1)2 = 1.

So, LHS = RHS.

It is proved that P (1) is true.

Now in second step by using induction hypothesis of mathematical induction we assume P (k) is true.

        1 + 2 + 3 + ……. + k = k(k+1)2

We need to prove P(k + 1) is true by using P (k) true.

For P(k + 1),

LHS = 1 + 2 + 3 + ……. + k + (k + 1)  

      = k(k+1)2 + (k+1) ………by using induction hypothesis

      = (k+1)(k+2)2

      = (k+1)((k+1)+12 = RHS for P(k + 1)

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

Thus P(1) is true and P(k + 1) is true whenever p(k) is true.  

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

Mathematical Induction - Problems with Solutions (induction proof):

1. Using the principle of mathematical induction, prove that n(n + 1)(n + 5) is a multiple of 3 for all n ∈ N.  

Solution:  

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

For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.  

So, the given statement is true for n = 1, i.e. P(1) is true.  

Let P(k) be true. Then,  

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

⇒ K(k + 1)(k + 5) = 3m for some natural number m, ... (i)  

Now, (k + 1l)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2)  

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

                            = k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2)  

                            = k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2)  

                            = k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification]  

                            = 3m + 3(k + 1 )(k + 4) [using (i)]  

                            = 3[m + (k + 1)(k + 4)], which is a multiple of 3

⇒ P(k + 1): (k + 1 )(k + 2)(k + 6) is a multiple of 3

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

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.  

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

Answer 2

$\huge\mathfrak{Hola\: Mate}$

$\huge\bold{Solution}$

Let P(n) be the statement "n² + n is even" .

We have , P(n) : n² + n is even

•.• 1² + 1 = 2 , which is even

.•. P(1) is true

Let P(m) be true . Then,

P(m) is true => m² + m is even => m² + m = 2lambda

for some lambda € N

Now , we shall show that P(m + 1) is true . for this we have to show that (m + 1)² + (m + 1) is an even natural number.

Now,

(m + 1)² + (m + 1) = (m² + 2m + 1) + (m + 1) = (m² + m) + (2m + 2)

=> (m + 1)² + (m + 1) = m² + m + 2(m + 1) = 2lambda + 2(m + 1)

=> (m + 1)² + (m + 1) = 2( lambda + m + 1) = 2ù

, where ù = lambda + m + 1 € N.

=> (m + 1)² + (m + 1) is an even natural number.

=> P(m + 1) is true

Thus , P(m) is true => P(m + 1) is true

Hence,

By the principle of mathematical induction, P(n) is true for all n € N that is n² + n is even for all n € N.

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