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