Explicație pas cu pas:
suma se poate scrie:
[tex]r = \frac{1}{8} + \frac{1}{24} + \frac{1}{48} + \frac{1}{80} + ... + \frac{1}{9800} = \\ = \frac{1}{ {3}^{2} - 1} + \frac{1}{ {5}^{2} - 1} + \frac{1}{ {7}^{2} - 1} + \frac{1}{ {9}^{2} - 1} + ... + \frac{1}{ {99}^{2} - 1}[/tex]
a) din recurența șirului, deducem că avem:
(99 - 3)÷2 + 1 = 49 termeni
b)
[tex]r = \frac{1}{4}\left(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ... + \frac{1}{2450}\right) = \\ [/tex]
[tex]= \frac{1}{4}\left(\frac{1}{1\times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + ... + \frac{1}{49 \times 50} \right) \\ [/tex]
[tex]= \frac{1}{4}\left( \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{49} - \frac{1}{50}\right) \\ [/tex]
[tex]= \frac{1}{4}\left( \frac{1}{1} - \frac{1}{50}\right) = \frac{1}{4}\cdot \frac{49}{50} = \frac{1}{4}\cdot \frac{98}{100} \\[/tex]
[tex]0.24 = \frac{24}{100} = \frac{96}{400} < \frac{98}{400} \\ [/tex]
[tex]0.25 = \frac{100}{400} > \frac{98}{400} \\ [/tex]
[tex]\iff 0.24 < \frac{98}{400} < 0.25 \implies 0.24 < r < 0.25 \\ [/tex]
