Prove the Pythagoras theorem using calculus, by using the fact that the area of a circle is proportional to the square of its radius. I was reading a routine morning message in a local chat group. There were some interesting facts in the message. One such fact was that we can prove the Pythagor...

Evaluate the following series $$\sum_{n=0}^{2023}\frac{2^n}{1+2023^{2n}}$$ My first thoughts were that this series didn't seem to telescope so there must be some other way of solving it. I did $$S=\sum_{n=0}^{2023}\frac{2^n}{1+2023^{2n}}$$ $$\implies S=\sum_{n=0}^{2023}\frac{2^{2023-n}}{1+2023^...

I often want to improve the tags on questions like "Is it possible to make this expression shorter?". This question asks how to simplify or evaluate the finite sum $$ \sum_{i=1}^n\left( 2^{i-1} a_i \sum_{k=0}^{n-i}10^k \binom{n-i}{k} \right)$$ Questions of this type seem to be fairly common,...

Evaluate $$\sum_{n=1}^{119}\sqrt{2-\frac{1}{n^2+\sqrt{n^4+\frac{1}{4}}}}$$ I first decided to do some simplifications. I rewrote $n^4+\frac14$ as $\left(n^2+n+\frac12\right)\left(n^2-n+\frac12\right)$ But I cannot get things out of the square root. Any hints or solutions are greatly appreciated.

I have these subsets of $R^n$. $A=\{B(x,r): x \in{R^n}, r \in Q^+\}$ $B=\{B(x,r): x \in{Q^n}, r \in Q^+\}$ $C=\{B(x,r): x \in{R^n}, r > 1 \}$ how can i prove are basis for the euclidean topology?

I have these subsets of $R^n$. $A=\{B(x,r): x \in{R^n}, r \in Q^+\}$ $B=\{B(x,r): x \in{Q^n}, r \in Q^+\}$ $C=\{B(x,r): x \in{R^n}, r > 1 \}$ how can i prove are basis for the euclidean topology?