Question

n²>2n+1 for all n>=3
solve using induction method​

Answer 1

Question :-

Prove by using Induction method

$\rm \: {n}^{2} > 2n + 1 \: \: \forall \: n \geqslant 3$

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: P(n) : {n}^{2} > 2n + 1 \: \: \forall \: n \geqslant 3$

Step :- 1 For n = 3

$\rm \: P(3) : {3}^{2} > 3 \times 2 + 1 \: \:$

$\rm \: P(3) : 9 > 6 + 1 \: \:$

$\rm \: P(3) : 9 > 7 \: \:$

$\rm\implies \:P(n) \: is \: true \: for \: n \: = \: 3$

Step :- 2 Assume that P(n) is true for n = k, where k is natural number greater than or equals to 3

$\rm \: P(k) : {k}^{2} > 2k + 1 \: \: \forall \: k \geqslant 3 - - - - (1)$

Step :- 3 We have to prove that P(n) is true for n = k + 1.

$\rm \: P(k + 1) : {(k + 1)}^{2} > 2k + 3 \: \: \forall \: k \geqslant 3 - - - - (2)$

Now, from step 2, we have

$\rm \: {k}^{2} > 2k + 1$

On adding 2k + 1 on both sides, we get

$\rm \: {k}^{2} + 2k + 1 > 2k + 1 + 2k + 1$

$\rm \: {(k + 1)}^{2} > 2k + 2 + 2k$

As,

$\rm \: k \geqslant 3\rm\implies \:2k \geqslant 6\rm\implies \:2k > 1$

So, using this,

$\rm \: {(k + 1)}^{2} > 2k + 2 + 2k > 2k + 2 + 1$

$\rm\implies \: {(k + 1)}^{2} > 2k + 3$

$\rm\implies \:P(n) \: is \: true \: for \: n \: = \: k + 1$

Hence, By the Process of Principal of Mathematical Induction,

$\: \: \: \: \: \: \: \: \: \boxed{\sf{ \: \:   \:\rm \: {n}^{2} > 2n + 1 \: \: \forall \: n \geqslant 3 \: \: \: \: }}$

Answer 2

★ Concept

Here the Principle of Mathematical Induction for proving inequality has been used. Note that for proving inequality by Induction method we have to go through with two steps- Basic Step and Induction Step.In this particular question where we have to prove an inequality true for all n (natural numbers) greater than equal to 3, in basic step we will prove that the inequality is true for n=3. In Induction step, first of all we have to assume that the inequality is true for n=k ; K is any natural number greater than 3. Secondly, we have to prove that the inequality is also true for n= k+1.

Let's proceed with proving !!

$\rule{190pt}{1pt}$

Let the mathematical statement be T(n)

T(n) = n² > 2n+1 ∀ n ≥ 3

☞ Basic Step

First of all we have to prove the equation true for n=3.

⇒ $\sf \: T(3) : {3}^{2} > 2(3) + 1$

➠ 9 > 7 , that's true.

∴ T(3) is true for above inequality.

☞ Induction Step

Step 1 : Assume that the inequality is true for n=k ; k is any natural number greater than 3.

⇒ $\sf \: T(k) : {k}^{2} > 2(k) + 1$

Step 2 : Now prove that the inequality is true for n=k+1

⇒ $\sf \: T(k + 1) : {(k + 1)}^{2} > 2(k + 1) + 1$

Now, From Step 1, we can write that

⇒ $\sf {k}^{2} > 2(k) + 1$

Adding 2k+1 on both sides, we get

⇒ $\sf {k}^{2} + 2k + 1> 2k+ 1 + 2k + 1$

Observe that k² + 2k + 1 is a formula for (k+1)².

⇒ $\sf {(k + 1)}^{2}> 2(k + 1) + 2k$

Here what we have to notice is that

⇒ $\sf 2(k+1) + 2k > 2(k+1) + 1.$

or we can write like

⇒ $\sf {(k + 1)}^{2}> 2(k + 1) + 2k > 2(k + 1) + 1$

So we have proved that

⇒ $\sf {(k + 1)}^{2} > 2(k + 1) + 1$

Conclusion

By using the principle of mathematical Induction we have proved that the given inequality is true ∀ n ≥ 3.

$\underline{\rule{190pt}{2pt}}$

☞ More to know

In Basic Step we usually prove T(n) true for n=1 but here in the question we have to prove that the inequality holds true for all those natural numbers greater than equal to 3 so we cannot take n=1 in Basic Step.