Understanding Summation Series: Finding Patterns in Sums of Squares

[jobsism]

Sum of 1^2 +2^2 +...k^2

How do you evaluate the sum of the series:-

(1^2) + (2^2)+...(k^2) ?

I do know how to prove that the sum is k(k+1)(2k+1)/6 by induction, but I'm just curious: how do you figure out the pattern for this, and other similar summation series?

 

[micromass]

Interesting information:

homepages.math.uic.edu/~kauffman/DCalc.pdf

and

www.math.upenn.edu/~wilf/gfologyLinked2.pdf

 

[Tobias Funke]

I like this geometric proof.

http://takayaiwamoto.com/Sums_and_Series/sumsqr_1_anim.gif

 

[dodo]

homepages.math.uic.edu/~kauffman/DCalc.pdf

Hey, thanks a lot! This is amazing. (The second link will take more time to digest, of course.)

If you are the author, you may want to fix a typo (I think) at the bottom of page 3, where $12n^{\underline 3}$ becomes $8n(n-1)(n-2)$ instead of $12n(n-1)(n-2)$.

 

[jobsism]

Thanks you, micromass and Tobias Funke! :D

I didn't even know that there was a branch of mathematics called generatingfunctionology! I didn't read much of it, but I intend to. The article on Discrete Calculus was more interesting, but it's taking some time to get used to it. Can you suggest some introductory books for it, to gain practice?

That geometric proof was simply marvellous, but such proofs can't exist for more sophisticated series, can it?

 

[scurty]

That geometric proof was simply marvellous, but such proofs can't exist for more sophisticated series, can it?

There are these series of books out there labeled Proof Without Words by Roger B. Nelsen which are FILLED with similar examples to this type of problem. They don't provide the proof, that's left up to you, but all of the content is viewed geometrically.

I've attached three "proofs"from the book that I thought were interesting.

 

[jobsism]

Those proofs are amazing! Thanks, scurty! I'll check out that book.

The third one reminded me of the Sierpinski fractal. :D

 

[mathwonk]

this is explained in a footnote on page 27 of courant's calculus.