How to derive $\sum j^2$ from telescoping property

Question

The book Real Analysis via Sequences and Series has a method of proving that $$\sum_{j=1}^n j = \frac{n(n+1)}{2}$$ that I've never seen before. The way they do it is by starting with $\sum (2j+1)$, using the fact that $2j+1 = (j+1)^2-j^2$, and then using the telescoping property.

I find this method very aesthetically pleasing, but I have two questions about this: $(1)$ what was the motivation for starting with $2j+1$? Why would that have come into the authors' minds as a way of deriving a formula for $\sum j$? And, a related question: $(2)$, after that derivation the authors state that the formula for $\sum_{j=1}^n j^2$ can be found in a similar way. I haven't been able to figure out which telescoping series I should equate this to. How can the formula for this summation be found similarly?

Answer 1

Based on the previous method it seems plausible to start with the fact that $(j + 1)^3 - j^3 = 3j^2 + 3j + 1$ Summing both sides gives:

$\sum\limits_{j=1}^n (j+1)^3 - j^3 = 3\sum\limits_{j=1}^n j^2 + \sum\limits_{j=1}^n 3j + \sum\limits_{j=1}^n 1$

The left hand side telescopes, so we get

$(n+1)^3 - 1 = 3\sum\limits_{j=1}^n j^2 + \frac{3n(n+1)}{2} + n$

So:

$\sum\limits_{n=1}^n j^2 = \frac{(2n^3 + 6n^2 + 6n) - (3n^2 + 3n) - (2n)}{6} = \frac{2n^3 + 3n^2 + n}{6} = \frac{n(n+1)(2n+1)}{6}$

Answer 2

The motivation for (1) might be the fact that many mathematicians just happen to know that summing the first few odd numbers gives the sequence of squares. The convincing proof without words is well known:

https://en.wikipedia.org/wiki/Proof_without_words

Here's a proof without words for the sums of squares:

http://mathandmultimedia.com/2012/05/27/proof-of-the-sum-of-square-numbers/

Answer 3

It's not so much 2j+1 that is natural to start with; it is more knowing the trick that if you do (j+1)^2 - j^2, then that will eliminate the quadratic part, reducing to a linear expression, which just happens to be 2j+1.