Question

Prove that 2^n > n where n is a natural number​

Answer 1

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

Given expression to prove is

$\rm :\longmapsto\: {2}^{n} > n$

where, n is a natural number.

We use Principal of Mathematical Induction to prove this statement.

Let assume that

$\rm :\longmapsto\: P(n) :{2}^{n} > n$

Step :- 1 For n = 1

$\rm :\longmapsto\: P(1) :{2}^{1} > 1\rm\implies \:2 > 1$

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

Step :- 2 Let assume that P(n) is true for n = k, where k is a natural number

$\rm :\longmapsto\: P(k) :{2}^{k} > k - - - (1)$

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

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

Now, from equation (1), we have

$\rm :\longmapsto\: {2}^{k} > k$

On multiply by 2 on both sides, we have

$\rm :\longmapsto\:2. {2}^{k} > 2k$

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

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

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

Hence, by the process of Principal of Mathematical Induction,

$\purple{\rm :\longmapsto\: {2}^{n} > n}$

Answer 2

Answer:

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

Given expression to prove is

$\rm :\longmapsto\: {2}^{n} > n$

where, n is a natural number.

We use Principal of Mathematical Induction to prove this statement.

Let assume that

$\rm :\longmapsto\: P(n) :{2}^{n} > n$

Step :- 1 For n = 1

$\rm :\longmapsto\: P(1) :{2}^{1} > 1\rm\implies \:2 > 1$

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

Step :- 2 Let assume that P(n) is true for n = k, where k is a natural number

$\rm :\longmapsto\: P(k) :{2}^{k} > k - - - (1)$

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

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

Now, from equation (1), we have

$\rm :\longmapsto\: {2}^{k} > k$

On multiply by 2 on both sides, we have

$\rm :\longmapsto\:2. {2}^{k} > 2k$

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

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

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

Hence, by the process of Principal of Mathematical Induction,

$\purple{\rm :\longmapsto\: {2}^{n} > n}$

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