Question

Rezolvare la ex din poza

Answer 1

$\displaystyle \textrm{Pentru }T_n, \textrm{ avem:} \\ \\ ip = \frac{1}{2^n}m = l\sqrt{2} \implies l = \frac{\sqrt{2}}{2\cdot 2^n}m\\ \\ p_n = \frac{1}{2^n}m + 2\cdot \frac{\sqrt{2}}{2\cdot 2^n}m = \frac{1}{2^n}m \cdot (1 + \sqrt{2})\\ \\ s_n = \frac{l\cdot l}{2} = \frac{2\frac{1}{2^n\cdot 2^n}\cdot m^2}{8} = \frac{m^2}{4\cdot 4^n}$

$\displaystyle S_n = \sum_{k=0}^{n} s_k= \sum_{k=0}^{n} \frac{m^2}{4\cdot 4^k} = \frac{m^2}{4}\sum_{k=0}^{n} \frac{1}{4^k} = \frac{m^2}{4}\cdot \frac{\frac{1}{4^{n+1}} - 1}{\frac{1}{4} - 1} = m^2 \cdot \frac{1 - \frac{1}{4^{n+1}}}{3}\\ \\ \lim\limits_{n\to \infty} S_n = m^2 \cdot \frac{1}{3} = \frac{m^2}{3}$

$\displaystyle P_n = \sum_{k=0}^n p_k = \sum_{k=0}^n \frac{1}{2^k}m \cdot (1 + \sqrt{2}) = m(1+\sqrt{2}) \sum_{k=0}^n \frac{1}{2^k} = m(1+\sqrt{2}) \frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}} = m(1+\sqrt{2}) \frac{1-\frac{1}{2^{n+1}}}{\frac{1}{2}} = 2m(1+\sqrt{2})\Big(1-\frac{1}{2^{n+1}}\Big)\\ \\ \lim\limits_{n\to \infty} P_n = 2m(1+\sqrt{2})$