Question

Show that (n^2) + 4 < (n + 1)^2 for all natural numbers n>=2

Answer 1

for \(n \geq 2\)
4 < 2n +1, right?

Answer 2

I plugged in 2 for n to show the base case.

Answer 3

base case n =2, 4 < 2n +1 =5, check
the left hand side is 4, a constant, never change. while the right hand side is increasing with n > 2

Answer 4

For example, if n =3 , 2n +1 = 7 >4
n =4, 2n +1 = 9 >4
and so on. We all have 2n +1 >4

Answer 5

This should be a mathematical induction proof.

Answer 6

no need

Answer 7

I mean it is required for my homework.

Answer 8

oh!! no need to use induction for this because it is so simple to get the proof

Answer 9

ok, I show you this first
4 < 2n +1
+n^2 both sides
you have n^2 +2 < n^2 +2n +1
the right hand side is (n+1)^2 done.

Answer 10

Now, if you want to use induction, ok, go ahead
basic case n =2
n^2 +4 = 8 < (n+1)^2 =(2+1)^2 =9 check

Answer 11

Suppose it holds for n = k , that is k^2 + 4 < (k+1)^2

Answer 12

Now prove it holds for n = k+1
that is (k+1)^2 +4 < ((k+1)+1)^2

Answer 13

Yeah, this is where I got stuck.

Answer 14

It might help to think of `k^2 + 4 < (k + 1)^2` as `k^2 + 4 + q = (k + 1)^2` where `q` is some positive number

Answer 15

ok, from k^2 +4 < k^2 +2k+1
add 2k +1 both sides
k^2 + 2k +1 +4 < k^2 + 2k +1 +2k +1
(k+1)^2 +4 < k^2 +4k +2
If we consider (k+2)^2 = k^2 + 4k + 4 which is greater than our left hand side

Answer 16

hence \((k+1)^2 +4< k^2 +4k +2 < k^2 +4k +4\)
pick far left and far right, we have what we need to prove. ok?

Answer 17

sorry for above, greater than our RIGHT hand side