Question

ajutati-ma pls cu exercitiile 7 si 8!!
dau coroana!!

Answer 1

Explicație pas cu pas:

7a) $\frac{1}{n+1} + \frac{2}{n+1} + ... + \frac{n}{n+1} = \frac{1+2+..+n}{n+1} = \frac{\frac{n(n+1)}{2}}{n+1} = \frac{n}{2}$

b) $\frac{2}{41} + \frac{4}{41} + ... + \frac{82}{41} = 2 * \frac{1+2+...+40}{41} + 2 = 2 * \frac{40}{2} + 2 = 42$

c)$\frac{4}{32}+ \frac{8}{32} + ... + \frac{400}{32}=\frac{4}{32}*( 1+2+..+100)=\frac{4}{32}*5050=\frac{5050}{8}=\frac{2525}{4}$

d)$\frac{10}{61} + \frac{20}{61} + ...+ \frac{600}{61} = \frac{10}{61} * (1+2 + ... + 60) = \frac{10}{61} * 30*61 = 300$

8a) $\frac{1}{2} + \frac{1}{2^2} + ... + \frac{1}{2^5} == \frac{16+8+4+2+1}{32} = \frac{31}{32}$

b)$\frac{1}{2}+\frac{1}{2^2} + ... + \frac{1}{2^{20}} = \frac{2^{20} - 1}{2^{20}}$

c)$\frac{1}{2}+\frac{1}{2^2} + ... + \frac{1}{2^{101}} = \frac{2^{101} - 1}{2^{101}}$

d)$\frac{1}{2}-\frac{1}{2^2} - ... - \frac{1}{2^5} = \frac{2^4}{2^5} - \frac{2^3}{2^5} - ... - \frac{1}{2^5} = \frac{16}{32} - \frac{8}{32} - \frac{4}{32} - \frac{2}{32} - \frac{1}{32}= \frac{16-8-4-2-1}{32} = \frac{1}{32}$