(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that the placement of the parentheses in any given sum is irrelevant.

2. Relevant equations

[tex]a + (b + c) = (a + b) + c[/tex]

[tex]a_1 + \dots + a_n = a_1 + (a_2 + \dots (a_{n-1} + a_n) \dots )[/tex]

3. The attempt at a solution

My book suggests that I split this problem up into three segments and I've followed that advice. I would appreciate it if someone could give me feedback on whatever of the following is correct and what is not correct. Thanks!

1. We first prove that

[tex](a_1 + \dots a_{n-1}) + a_n = a_1 + \dots a_n[/tex]

by induction on [itex]n[/itex]. Clearly eqaulity holds when [itex]n = 1[/itex], [itex]n = 2[/itex], and [itex]n = 3[/itex] (by the associative property). Now assume that this equality holds for the sum of [itex]k[/itex] numbers; if it also holds for [itex]k+1[/itex] numbers, then equality always holds. Consider,

[tex](a_1 + \dots + a_k) + a_{k+1}[/tex]

Using the associative property, this expression can be rewritten into the form

[tex](a_1 + \dots + a_k) + a_{k+1} = a_1 + [(a_2 + \dots a_k) + a_{k+1}][/tex]

Since this latter expression contains exactly [itex]k[/itex] numbers (in which case, equality holds) this sum can once again be rewritten as

[tex](a_1 + \dots + a_k) + a_{k+1} = a_1 + (a_2 + \dots a_k + a_{k+1}) = a_1 + a_{k+1}[/tex]

as desired.

2. Next, we prove that if [itex]n \geq k[/itex], then

[tex](a_1 + \dots a_k) + (a_{k+1} + \dots + a_n) = a_1 + \dots a_n[/tex]

by induction on [itex]k[/itex]. Clearly, equality holds when [itex]k = 1[/itex] since the expression on the left is immediately identical to the one on the right. Now suppose that equality holds for some arbitrary [itex]k[/itex]; if this implies that it also holds for [itex]k+1[/itex], then it equality always holds. So consider,

[tex](a_1 + \dots + a_{k+1}) + (a_{k+2} + \dots + a_n)[/tex]

By part 1 we know that

[tex]a_1 + \dots + a_{k+1} = (a_1 + \dots + a_k) + a_{k+1}[/tex]

In which case, we have the equality

[tex](a_1 + \dots + a_{k+1}) + (a_{k+2} + \dots + a_n) = [(a_1 + \dots a_k) + a_{k+1}] + (a_{k+2} + \dots + a_n)[/tex]

Using the associative property to rewrite this, we see that

[tex](a_1 + \dots + a_{k+1}) + (a_{k+2} + \dots + a_n) = (a_1 + \dots a_k) + [a_{k+1} + (a_{k+2} + \dots + a_n)] = a_1 + \dots + a_n[/tex]

By the assumption in our inductive step. This completes this part of the proof.

3. To complete the proof, suppose that [itex]s(a_1, \dots, a_n)[/itex] is some sum formed from [itex]a_1, \dots, a_n[/itex]. Then we need only show that

[tex]s(a_1, \dots, a_n) = a_1 + \dots a_n[/tex]

I figured that it might be possible to do this by utilizing the fact every sum can be split up such that [itex]s(a_1, \dots, a_n) = s'(a_1, \dots, a_k) + s''(a_{k+1}, \dots, a_n)[/itex]. If one were to continue to split each sum up, then one would invariably end up with a number of sums of the form [itex]a_l + \dots a_k[/itex]. Using part 2, the sum of these sums could then be shown to be equal to [itex]a_1 + \dots + a_n[/itex].

This was only my initial thought and the reasoning seems a bit shaky so I would appreciate any advice on how to complete this part of the proof. I would also appreciate corrections on parts 1 and 2 of the proof. Thanks!

Edit: I'm sorry if this is in the wrong section. It's a problem from my calculus book and it seems a bit more advanced than many pre-calculus problems so I figured that this would be the correct sub-forum.

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# Homework Help: Order of Parentheses in a Sum

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