Question

prove that 2^n>=n^2 for n>=4

Answer 1

use mathematical induction

Answer 2

i am at the point that 2^(k+1)>=(k+1)^2

Answer 3

than i multiplied by 2 the two sides

Answer 4

of this
2^k>=k^2

Answer 5

and i got 2^(k+1)>=2k^2

Answer 6

Start off with n = 4
2^4 = 4^2 GOOD
For n = 5
2^5 > 5^2 GOOD

Now assume that it holds for n, i.e.
2^n ≥ n^2
Let's try n+1
2^(n+1) = 2*2^n ≥ 2*n^2 = n^2 + n^2
Now since n ≥ 4,
n^2 ≥ 4n = 2n + 2n
and 2n > 1
So n^2 > 2n + 1
Thus 2n^2 > n^2 + 2n + 1 = (n+1)^2
Thus 2^(n+1) = 2*2^n > (n+1)^2

Answer 7

2^n >= n^2 for n >= 4

When n = 4, 2^n = 2^4 = 16; n^2 = 4^2 = 16. So 2^n >= n^2 is true.
When n = 5, 2^n = 2^5 = 32; n^2 = 5^2 = 25. So 2^n >= n^2 is true.

When n = k, 2^n = 2^k; n^2 = k^2.
Assume 2^k >= k^2.
With the assumption 2^k >= k^2 if we can prove that 2^(k+1) >= (k+1)^2 then we have proved the original question by induction.

If 2^k >= k^2, then:
2 * 2^k >= 2 * k^2
2^(k+1) >= 2k^2 ---- (1)

Now we need to prove 2k^2 >= (k+1)^2:
2k^2 >= k^2 + 2k + 1
k^2 - 2k >= 1
k^2 - 2k + 1 >= 2
(k-1)^2 >= 2
This is indeed true for all values of k > 2. The problem is only interested in k >= 4.

Therefore, 2k^2 >= (k+1)^2 for k >= 4. Put this in (1):
2^(k+1) >= (k+1)^2 for k >= 4

We started with the assumption that 2^k >= k^2 and proved that 2^(k+1) >= (k+1)^2. Therefore, by induction we have proved that 2^n >= n^2 for n >= 4.