# Spivak Chapter 2 problem 6

1. Apr 25, 2014

### Jef123

I'm having a hard time seeing how spivak derived this formula...

"6. The formula for 12+...+n2 may be derived as follows...

(k+1)3 - k3 = 3k2+3k+1 where k=1,...,n

Therefore, (n+1)3 - n3 = 3n2+3n+1

But this is where I am confused...He then presents this,

(n+1)3 - 1 = 3[12 +...n2]+3[1+...+n]+n

Where did the "1" come from on the left hand side of the equation? And where did the "n" come from at the far right hand side of the equation?

On a side note, I am having a lot of difficulty going through this text myself, mainly due to the fact that I feel a little lost with how to do proofs. I have done a calculus course before, but this text is much more difficult. Does anyone suppose that as I continue to go through the text that I will develop a better sense of how to do proofs and feel more comfortable with understanding this material? Any suggestions would be great

2. Apr 25, 2014

### eigenperson

This is the classic "telescoping" trick. Look again at the equation that includes k. If you plug in all the values for k in the given range, you get n equations. (In the book, the vertical "..." conceals most of the equations.) Then add all n of those equations together to get the last line (the one that is confusing you).

I'm pretty sure that if you try very hard to do all the problems in Spivak, you will soon find that you feel much more comfortable with the proofs. It would also be a good idea to show your solutions to someone else (preferably with more mathematical experience) to make sure you are on the right track.

3. Apr 25, 2014

### slider142

In case it wasn't clear, you first need to use the formula (k+1)3 - k3 = 3k2 + 3k + 1 with the n values k = 1, 2, 3, ..., n. This gives you a list of n equations. As always, if you are not sure what is going on, use an actual number for n, such as n = 5.
Add all n values on the left side of these n equations to get (n+1)3 - 1.
Most of the values on the left side are cancelled by their additive inverses. In particular, the first value, if you list the largest values first, is (n+1)3 - n3, while the second value is n3 - (n-1)3. Adding just these two values, we see that the n3 term disappears. Adding the third value, (n-1)3 - (n-2)3, makes the (n-1)3 term disappear, and so on until we come to the last term, where we have 23- 13. The 23 term is cancelled by the -23 in the previous term, so we are left with (n+1)3-1. This type of behavior is called a telescoping sum for short.
Since these values are equal to the values on the right side, the sum of all these values must equal the sum of all the values on the right side, which is 3[12 + ... + n2] + 3[1 + ... + n] + n.
I agree also to at least give serious effort to the problems in Spivak. Some of them are very difficult until you hit upon a particular way of thinking. The type of diligence necessary to solve problems that are not merely repeating a single problem-solving recipe several hundred times, as is present in other types of calculus texts, is very useful in many areas of life, not just mathematics. If you can't get one, skip it and go back to it another time, if necessary. Or ask mathematically inclined friends, professors, or on these forums.
In particular, you will find that Spivak's problems are usually a preamble to a concept in a later chapter, and some you will apply beyond this text if you go on to study real/complex analysis or differential geometry. Remember the general point that each problem is trying to make, and you will find many later problems much easier if you apply specific results of earlier problems.

Last edited: Apr 25, 2014
4. Apr 25, 2014

### Jef123

Thanks to both of you for your help, I get it know. To slider142, I have noticed that this is quite the departure from the math that I am accustomed to and I think as to what you mentioned, my real difficulty lies with the fact that its natural for me to use some sort of formula given to me and then apply it to a question. I'm just going through this text myself and plan to follow this one up with Calculus on manifolds or linear algebra done right by axler because I have yet to even touch that subject. Do you think that would be a good text to use for linear algebra or would there be a better one?

5. Apr 25, 2014

### slider142

Linear Algebra Done Right is an excellent text, as is Spivak's Calculus on Manifolds (this text actually applies several problems from Spivak's Calculus). However, they are both abstract texts: they emphasize the interconnection of various topics.
If you plan on applying these topics anywhere outside of pure higher mathematics, I recommend balancing their abstract treatment with some applications, as can be found in https://www.amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/0130414085/ref=dp_ob_title_bk, as well as any basic text on linear algebra that uses the traditional determinant-based approach (some applications really do rely more on specific types of matrices than properties of linear transformations).
If anything, these two texts will give you mental exercises a bit easier than the ones present in the flagship texts you cited, allowing a bit of relaxation and side exploration of particular details of application.

Last edited by a moderator: May 6, 2017