Excrcise 2. Let F, be the n-th Fibonacci number. Prove E = F„Fn+1- %3D k–1

. Let $n \in \mathrm{N}^{*}$ and the equation:$$C_{2 n+1}^{1} x^{n}-C_{2 n+1}^{3} x^{n-1}+C_{2 n+1}^{5} x^{n-2}-\ldots+(-1)^{k} C_{2 n+1}^{2 k+1} x^{n-k}+\ldots+(-1)^{n} C_{2 n+1}^{2 n+1}=0$$Show that:a) the equation admits as roots the numbers::$\mathrm{x}_{\mathrm{k}}=\operatorname{ctg}^{2} \frac{k \pi}{2 n+1}$, for every $k \in\{1.2, \ldots . n\}$  b) $\frac{\pi^{2} n(2 n-1)}{3(2 n+1)^{2}}<\sum_{k=1}^{n} \frac{1}{k^{2}}<\frac{2 \pi^{2} n(n+1)}{3(2 n+1)^{2}}$ c) $\frac{\pi^{4} n(2 n-1)\left(4 n^{2}+10 n-9\right)}{45(2 n+1)^{4}}<\sum_{k=1}^{n} \frac{1}{k^{4}}<\frac{8 \pi^{4} n(n+1)\left(n^{2}+n+3\right)}{45(2 n+1)^{4}}$  d) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}$  e) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k^{4}}=\frac{\pi^{4}}{4}$  Please check the attached picture for details. I need complete solution with explanation please. the all fifth subpoints are for the same problem and they are conected between them, but if…

п(п+1)(п+2) E}=1jG+ 1) = а. b. Prove that the sum of the first n Fibonacci numbers with odd indices is given by: F, + F3 + F5 + …. + F2n-1 = F2n

Find the first natural number m such that 3^m > 8m^2. Show the details of your search.