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I mean, you have to show why that sum is \frac{\pi^2}{12}, but there are many ways of doing that in 4U, so it's basically done. That was also my method :).

In general, the number of ways of choosing a cards from b consecutive cards such that there are at least c cards between any pair of cards is \binom{b+c-ac}{a}, which isn't too difficult to prove using double-counting.

We know that by \frac{\text{d}}{\text{d}n} (f(n))^a = a f'(n) (f(n))^{a-1}, \frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1} We also know that by \frac{\text{d}}{\text{d}n} a^n = a^n \ln(a)...

Another problem if u get bored of that one: \displaystyle \int_0^1 \frac{\ln(x+1)}{x} \; \mathrm{d}x = \frac{\pi^2}{12} No using dilogarithms, only 4U stuff.

inb4 unprovable in ZFC

First q: Note that T_n T_{n+2} - T_{n+1}^2 = T_n^2 + T_n T_{n+1} - T_n^2 - 2T_n T_{n-1} - T_{n-1}^2 = T_n (T_{n+1} - 2T_{n-1}) - T_{n-1}^2 = T_n T_{n-2} - T_{n-1}^2, and then check base cases to get the result. Second q: Note that (a+b)^{2n+1} - a^{2n+1} - b^{2n+1} = (a+b)^2[(a+b)^{2n-1} -...

Anyone manage to solve Q11a?

We construct K such that CFK is an equilateral triangle with those points going around the triangle anti-clockwise. Then, we note that CB=CA, CK=CF and \angle BCK = \angle BCA - \angle KCA = 60^{\circ} - \angle KCA = \angle KCF - \angle KCA = \angle ACF (we're using directed angles here)...

16a was proving the Dandelin spheres property 16b was proving the eccentricity of an ellipse is equal to the ratio of the sines of the angle the plane cuts through the cone to the sine of the angle of the cone 16c was determining the probability a game (in which players A and B flip a coin till...

That's what I like about this question - due to the simplicity of the q, there seem to be so many ways of approaching it which don't go anywhere

In general, the number of ways of choosing a cards from b consecutive cards such that there are at least c cards between any pair of cards is \binom{b+c-ac}{a}, which isn't too difficult to prove using double-counting.

That's an excellent variety of integrals, none of which are on the right track (from what I've explored at least). My method is not that nice but it does yield a solution. My first hint is to try x=\tan^{-1}(u) in the original integral.

He says its a unit circle. Of course, for a circle of radius R, the answer is just \frac{4R}{\pi}.

You'll figure it out, I believe in you :)

I believe its \frac{4}{\pi}?

Let's make it more fun by using complex numbers lol (and here's a crash course of doing geometry with complex numbers). We can assume the circle is the unit circle, and O=0 is the origin of the Argand Plane. Also, let A=a, B=b, C=c, D=d such that |a|=|b|=1. Then, we can assume c=ka and d=lb...

no, unfortunately, i simply didnt have consistent enough performance :/

How so?

\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} (\ln(1+\sqrt{2}))^2

Pretty easy by the following: \frac{\sin(x)}{\sin(x)+\cos(x)} = \frac{1}{2} \left (1 - \frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)} \right )