show that (2^n)>n^2 whenever n is an integer greater than 4 Solution 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.

1. show that (2^n)>n^2 whenever n is an integer greater than 4
Solution
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