Mathematical Induction Question

Let the proposition P(n) be 2^n > n^2 for n > 4.
1. Consider P(5),
L.H.S.
= 2 ^ 5 = 32
R.H.S.
= 5 ^ 2 = 25
L.H.S. > R.H.S.
So P(5) is true.

2. Assume that P(k) is true.
i.e. 2^k > k^2 for k > 4

3. Consider P(k + 1),
L.H.S.
= 2^(k + 1)
= 2 * 2^k
> 2 * k^2
= k^2 + k^2
> k^2 + 4k [because k > 4, k^2 > 4k where k is positive]
= k^2 + 2k + 2k
> k^2 + 2k + 1 [because k > 4, 2k > 8 > 1]
= (k + 1)^2
L.H.S. > R.H.S.
So P(k + 1) is also true if P(k) is true.

Therefore, the proposition is true for n > 4.

Sub n=5
>5²
32>25
∴2^n＞n² when n＞4.

sub n=k
2^k>k² when n＞4.

snb n=k+1
2^(k+1)>(k+1)²
2^k × 2 >k²+2k+1
...........

Base Case: when n = 5, lhs = 2^5 = 32 and rhs = 5^2 = 25, => lhs > rhs. The statement holds when n = 5.

Assume the statement is true for n = k, i.e. 2^k > k^2.
For n=k+1, 2^(k+1) - (k+1)^2 = 2x2^k - (k^2 + 2k + 1) > k^2 - 2k - 1 = k^2 - 2k + 1 - 2
= (k - 1)^2 - 2
> (4 - 1)^2 - 2
= 7 > 0 for k > 4
The statement is also true for n = k + 1.

Therefore the statement is true for all n > 4.