Colin Beveridge @icecolbeveridge@mathstodon.xyz

Let \(S = \sum_{r=1}^n r^2\).

Then \(S = \\
1+\\
2+2+\\
3+3+3+\\
\cdots \\
n+n+n+\cdots+n\)

Also, \(S = \\
n+\\
n+(n-1)+\\
n+(n-1)+(n-2)+\\
\cdots \\
n+(n-1)+(n-2)+\cdots+2+1\)

And \(S = \\
n+\\
(n-1)+n+\\
(n-2)+(n-1)+n+\\
\cdots \\
1+2+3+\cdots+(n-1)+n\)

So \(S+S+S = \\
(2n+1)+\\
(2n+1)+(2n+1)+\\
(2n+1)+(2n+1)+(2n+1)+\\
\cdots \\
(2n+1)+(2n+1)+(2n+1)+\cdots+(2n+1)+(2n+1)\)

i.e., \(3S = (2n+1) \times \frac{n(n+1)}{2}\) or \(S = \frac{1}{6} n(n+1)(2n+1)\) #proofinatoot (via Jeremy Kun)

But for ellipses
We'd have no eclipses
#groot
#reckonthatmustbetrue

A post that features some of my Mathstodon tooters: http://www.flyingcoloursmaths.co.uk/summary-summery-summation/

Is \( \sum_{r=100}^{0} r \):

a) 5050
b) 0
c) -5050
d) other

I have an Opinion, but I don't want to share it until I know whether it's controversial (or worse yet, Wrong.)

No, wait, got it, it's in the column settings.

Is there an easy way to turn off the 'bloop' notifications? I can't see it in the settings.

Proof: Pythagoras.

Let \(C_1\) and \(C_2\) be circles that intersect at right angles, having \(P_1\) and \(P_2\) as their centres. Then the circle with diameter \( \vec{P_1 P_2} \) has a quarter of the area of the sum of the areas of \( C_1 \) and \( C_2 \).