# Prove by induction

1. Nov 9, 2008

### mamma_mia66

1. The problem statement, all variables and given/known data
Prove by induction that n2>(4n+7) for all integers n$$\geq$$6.
[Hint: somewhere you may use the fact that (2k+1)$$\geq$$13 > 4.]

2. Relevant equations

3. The attempt at a solution

n2> (4n+7)

(B) when n =6
62> (4*6+7)
36>31 true

(I) given n=k then k2> (4k+7)

let n=k+1= k2+2k+1

(k2+ 2k+1) > 4k+7 +2k+1

4k+7+2k+1> 4k+7+4

6k+8 > 4(k+1)+7

2. Nov 9, 2008

### Staff: Mentor

It's hard to follow the logic in what you wrote.

For the step you're trying to prove, you should have something like this:
(k + 1)^2 > ... > 4(k + 1) + 7
In other words, your inequality starts off with (k + 1)^2 and ends with a smaller quantity, 4(k + 1) + 7.

3. Nov 10, 2008

### mamma_mia66

Okay. Thank you.

4. Nov 10, 2008

### HallsofIvy

Staff Emeritus
Surely this is not what you meant to write! you meant, I think, that "let n= k+1, then n2= (k+1)2= k2+ 2k+ 1

[/quote](k2+ 2k+1) > 4k+7 +2k+1

4k+7+2k+1> 4k+7+4 [/quote]
It would be useful to point out that because k> 5, 2k+1> 11> 4