Proving \sum_{i=0}^n (i-3) \geq \frac{n^2}{4} with Mathematical Induction

[James889]

Hi,

I need some help with mathematical induction

The question is as follows:

prove that $$\sum_{i=0}^n (i-3) \geq \frac{n^2}{4}$$

I have shown that it holds for the base step where n=12
$$\frac{144}{4} = 36$$
and the sum of all the i's up to 12 $$-3,-2,-1,0,+1,+2,+3,+4,+5,+6,+7,+8,+9 = 39$$

$$39\geq36$$

Now for the inductive step:
$$\sum_{i=0}^{n+1}$$
and i know this can be rewritten, but I am not sure how
either it's $$\sum_{i=0}^{n} (i-3)(i-3)~~ \text{or}~~\sum_{i=0}^{n} (i-3)+(i-3)$$

How do i proceed from here?

 

[lanedance]

i think you need
$$\sum_{i=0}^{n+1} (i-3) = ((n+1)-3) + \sum_{i=0}^{n} (i-3)$$

then assuming the proposition is true for n leads to
$$\sum_{i=0}^{n+1} (i-3) = ((n+1)-3) + \sum_{i=0}^{n} (i-3) \geq ((n+1)-3) + \frac{n^2}{4}$$

 

[HallsofIvy]

The last term, that you want to take out of the sum, has i= n+1 so i- 3= n+1-3= n- 2.

$$\sum_{i=0}^{n+1}(i- 3)= \sum_{i=0}^n (i- 3)+ (n+1- 3)= \sum_{i=0}^n (i-3)+ n- 2$$

 

[James889]

The last term, that you want to take out of the sum, has i= n+1 so i- 3= n+1-3= n- 2.

$$\sum_{i=0}^{n+1}(i- 3)= \sum_{i=0}^n (i- 3)+ (n+1- 3)= \sum_{i=0}^n (i-3)+ n- 2$$

Okay,
Then you have to add (n+1) to the other side of the equation as well?
$$\sum_{i=0}^n (i-3)+ n- 2 = \frac{(n+1)^2}{4} + (n+1)$$

 

[HallsofIvy]

Okay,
Then you have to add (n+1) to the other side of the equation as well?
$$\sum_{i=0}^n (i-3)+ n- 2 = \frac{(n+1)^2}{4} + (n+1)$$

?? No, add n- 2 to other side as well!

 

[James889]

?? No, add n- 2 to other side as well!

Haha, I am so bad at this, it's almost funny