Mathematical Induction am I on the right track?

[tangibleLime]

Homework Statement

The sequence a_{0 -> n} is defined by a_i = b+i*c. Prove by induction on n that the sum of the terms in the sequence is (n+1)(a₀ + a_n)/2.

Homework Equations

The Attempt at a Solution

I defined predicate P(n) as (n+1)(a₀+a_n)/2.

My goal is P(n+1), which is (n+2)(a₀+a_n+1)/2. I believe I may have made a mistake here. For some reason I think it might be (n+2)(a₀+a_n+a_n+1)/2, but I'm not sure.

So to prove P(n+1), I take P(n) and add the new term to it, (a_n+1).
P(n) + a_n+1
= ((n+1)(a₀+a_n)/2) + (a_n+1)
= ((n+1)(a₀+a_n) + 2(a_n+1))/2

Obviously what I got here is not visually equal to P(n+1), but maybe I'm missing something simplifying-wise? Or did I make a mistake somewhere?

And yes, I know I also have to test the base case, which I can do easily.

Any help would be appreciated!

 

[HallsofIvy]

Homework Statement

The sequence a_{0 -> n} is defined by a_i = b+i*c. Prove by induction on n that the sum of the terms in the sequence is (n+1)(a₀ + a_n)/2.

Homework Equations

The Attempt at a Solution

I defined predicate P(n) as (n+1)(a₀+a_n)/2.

Well, this itself is incorrect. The "predicate" has to be a statement or an equation, not just an expression. Your predicate should be "The sum of the n terms in the sequence is [itex](n+1)(a_0+ a_n)/2[/itex].

My goal is P(n+1), which is (n+2)(a₀+a_n+1)/2. I believe I may have made a mistake here. For some reason I think it might be (n+2)(a₀+a_n+a_n+1)/2, but I'm not sure.

No, you were right the first time- although, again, P(n+1) should be the statement that the sum of n+1 terms is equal to that.

So to prove P(n+1), I take P(n) and add the new term to it, (a_n+1).
P(n) + a_n+1
= ((n+1)(a₀+a_n)/2) + (a_n+1)
= ((n+1)(a₀+a_n) + 2(a_n+1))/2

Obviously what I got here is not visually equal to P(n+1), but maybe I'm missing something simplifying-wise? Or did I make a mistake somewhere?

You haven't used the fact that [itex]a_i= b+ ic[/itex]!

Your predicate really is
[tex]\sum_{i=0}^n b+ ic= (n+1)\frac{b+ (b+ nc)}{2}= (n+1)\frac{2b+ nc}{2}[/tex]

That's what you really want to prove.

And yes, I know I also have to test the base case, which I can do easily.

Any help would be appreciated!