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