Using Telescoping Property for Summing ∑(2k-1)

[unintuit]

1+3+5+...+(2n-1)=∑(2k-1)

but (2k-1)=k²-(k-1)²
summing we use the telescoping property and deduce that ∑(2k-1)=n²-0²=n²

This seems accurate to me. Now my question is this a proper use of the telescoping property. In the least it reveals the proper answer, which can then be proved by induction.

 

[Stephen Tashi]

You example illustrates the use of a telescoping sum to reach a correct conclusion about finite sums. This is a reliable method of reasoning about finite sums. Is that what you mean by the words "accurate" and "proper" ?

The problem of finding simple expressions for finite sums can be regarded as the problem of "anti-differencing", which has an analogy to antidifferentiation in calculus. You can find material online about doing anti-differencing and there are rules for it that are analogous to those used for antidifferentiation. For example, there is Summation By Parts ( http://en.wikipedia.org/wiki/Summation_by_parts ) which is analgous to Integration By Parts.

 

[unintuit]

Yes to your first question. I was also interested in how one would determine the limit of the expression Σ[k²-(k-1)²]=Σ(2k-1). How would one know if it were to approach 0 vs. 1.
I assumed that it went to zero because
i. (k+1)²=k²+2k+1=(k+1)²-k²=2k+1 then summing over n we have [(n+1)²-(1+n)]/2=∑k. subtracting 1 from the terms in i. gave [k²-(k-1)²]→(n+1-1)²-(1-1)²=n². I did not know how to determine the limit which is why I was curious if this is a good way to figure out the sum vs. some unknown way. If there is some other way, what would it be?

 

[Stephen Tashi]

Yes to your first question. I was also interested in how one would determine the limit of the expression Σ[k²-(k-1)²]=Σ(2k-1).

What limit are you talking about? The limit of the expression as k approaches infinity? - or the limit of the expression as k approaches zero? - or as k approaches some other number?

 

[unintuit]

Nevermind, I figured out my mistake. I was thinking about it in the wrong way. It was starting with k=1 and k_i∈ℕ with k₁<k₂.
Thank you for your help.