Calculating Sum, intro to integration

[quantum_enhan]

Homework Statement

Can someone please explain the following for me? \sum ^{n}_{k=m+1} k^2 = \sum ^{n}_{k=1} k^2 - \sum ^{m}_{k=1} k^2<br />

The Attempt at a Solution

I have not done math in a long time, and for some reason this has me baffled. My textbook randomly does it, and does not explain how it got it. Thanks in advance

 

[micromass]

Well, what is

(1^2+2^2+3^3+4^2+...+n^2)-(1^2+2^2+3^3+4^3+...+m^2)

As you can see: a lot of terms will vanish. Thus the 1^2 will vanish, for example. But what terms will remain??

 

[quantum_enhan]

n^2-m^2. However, let's say you only have this term \sum ^{n}_{k=m+1} k^2, how would you know to simplify it to n^2-m^2?

 

[gb7nash]

n^2-m^2. However, let's say you only have this term \sum ^{n}_{k=m+1} k^2, how would you know to simplify it to n^2-m^2?

The RHS of the equation does not simplify to n^2-m^2. After the terms cancel out, we should be left with (m+1)^2 + ... + n^2

Unless I'm misunderstanding what you're asking.

 

[micromass]

Hmm, let's make it more concrete.

What is

(1^2+2^2+3^3+4^2+5^2)-(1^2+2^2)

That is not n^2-m^2, is it?

 

[quantum_enhan]

I guess I am confused as to why they randomly changed the lower limit from k=1+m to k=1, and how they did it

 

[ystael]

Forgive me if this explanation is too elementary, but you seem to be having trouble with what the notation means. First of all, there is an unwritten assumption here that 1 \leq m \leq n, so all that follows will suppose that.

Think of the symbol \sum_{k=1}^n k^2 as an instruction to add up the values in n labeled boxes. The boxes are labeled 1 through n, and each box contains the value which is the square of its label: if the box is labeled k, the value in it is k^2.

Now look at the right-hand side. \sum_{k=1}^n k^2 - \sum_{k=1}^m k^2 means: take the sum of the values in n red boxes labeled 1 through n, where the value in red box k is k^2, and subtract the sum of the values in m blue boxes labeled 1 through m, where the value in blue box k is k^2.

At this point we notice that the rule for putting values in the red boxes and the rule for putting values in the blue boxes are the same, so we can cancel a red box by subtracting off the corresponding blue box. We are left only with the red boxes that didn't have a corresponding blue box, that is, those with labels that are not between 1 and m. What are the remaining labels? The ones between m+1 and n. Thus the result of subtracting is \sum_{k=m+1}^n k^2.

Incidentally, this value is not n^2 - m^2. You can compute it without too much trouble (proving the formula for the sum of the first n squares is a standard exercise in mathematical induction). But you should be able to see immediately that n^2 - m^2 is wrong, because you are adding up a bunch of positive terms the largest of which is n^2, so you shouldn't be able to get something smaller than n^2.

(Why are some boxes red and some boxes blue? To make the point that the terms of different \sum signs need not have anything to do with each other. There's no clever cancellation trick you can do with \sum_{k=1}^n k^2 - \sum_{k=1}^m \log k.)

 

[quantum_enhan]

Thank you. Do you have any suggestions as to how one would approach the same problem, but going the other way? For example, evaluate: \sum ^{n}_{k=m+1} k^2. Thanks for your help

 

[ystael]

The underlying point is that actually the easiest way to evaluate \sum_{k=m+1}^n k^2 is to break it up into the difference shown. This is an example of a standard trick for putting expressions in convenient forms: you add zero by adding and subtracting the same thing, or multiply by one by multiplying and dividing by the same (non-zero) thing. In this case,

\sum_{k=m+1}^n k^2 = \sum_{k=m+1}^n k^2 + \left(\sum_{k=1}^m k^2 - \sum_{k=1}^m k^2\right) = \left(\sum_{k=m+1}^n k^2 + \sum_{k=1}^m k^2\right) - \sum_{k=1}^n k^2 = \sum_{k=1}^n k^2 - \sum_{k=1}^m k^2.

In the center equation, the only difference between the two sides is the parentheses; this is an application of the associative law for addition.

(I edited my previous post since your reply to clarify another point or two, so you may wish to look at it again.)