Im not good at induction involving inequalities; can someone help me prove by mathematical induction:

1) 2^n - n^2 > 0 where n>4

2) n^2 - 3n + 2 ≥ 0

3) 4^n - 1 - 7n ≥ 0 where n ≥2

thanks

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- Mar 17th 2009, 12:45 AMnoobonastickMathematical Induction [inequalities]
Im not good at induction involving inequalities; can someone help me prove by mathematical induction:

1) 2^n - n^2 > 0 where n>4

2) n^2 - 3n + 2 ≥ 0

3) 4^n - 1 - 7n ≥ 0 where n ≥2

thanks

- Mar 17th 2009, 01:34 AMADARSH
All the questions are almost same try them after seeing this

P(n) :2^n - n^2 >0 When n>4

**First step:**

Now 7 >0 hence its correct for n = 5

**Assumption :**P(k) is correct

ie;

Hence

**Proof for P(k+1) using assumption:**

since

Hence

But since (k+1)^2 is always a positive number (greater than 16 atleast ) thus proved that its correct for k+1 hence through the principle of mathematical induction this is proved - Mar 17th 2009, 02:00 AMnoobonastick
I can't seem to get the answer to the last question.

I have to prove 21k - 4 > 0 ... but how do I do that.

Do I state that if I sub in k ≥ 2, then 21k - 4> 0? - Mar 17th 2009, 03:35 AMADARSH
- Apr 8th 2010, 12:17 PMdaniel
I know the rules state create a new thread for each problem, but this isn't really a new problem. Rather, it's a question regarding the solution posted, as I'm unsure as well about the application of induction to inequalities.

Reading the Assumption step, I'm not sure how that would work if p(n) = 2^n > n^2, for n>4. Would it essentially be the opposite to what was posted, becoming "hence: 2^n - n^2 > 0" ? - Apr 8th 2010, 12:40 PMArchie Meade

is the same as

Here is another way to prove it by induction..

**P(k)**

for k>4

**P(k+1)**

Try to prove that

causes

**Proof**

?

?

?

If k>4, then

hence, if

then

- Apr 8th 2010, 12:58 PMArchie Meade
For the 3rd..

**P(k)**

?

?

**P(k+1)**

Does cause

?

**Proof**

?

?

?

If

then is definately greater than 7.

Finally show P(k) it is valid for k=2 - Apr 9th 2010, 07:35 AMnovice
When we prove by Principle of Mathematical Induction, as much as it is possible, we want it be more convincing. Since the theorem says

It is more convincing to begin with and end with . By beginning with P(k+1) is less desirable though it is not always the case for Strong Principle of Mathematical Induction.

As much as we could, we would begin with

.

Now subtract both sides by , and we arrive at

, which is precisely as desired; namely, P(k+1).

.

Notice that the first line of proof was

and the last line of proof is .

That is to say we have proved

for