Răspuns:
progresie geometrică
Explicație:
[tex]S_{n} = \frac{b_{1}\cdot ( {q}^{n} - 1)}{q - 1} \\ [/tex]
b)
[tex]b_{1} = 7, q = 7, n = 43[/tex]
[tex]b = 7 + {7}^{2} + {7}^{3} + ... + {7}^{43} = \\ = \frac{7\cdot ( {7}^{43} - 1)}{7 - 1} = \frac{7\cdot ( {7}^{43} - 1)}{6}[/tex]
c)
[tex]b_{1} = 1, q = 3, n = 43[/tex]
[tex]c = {3}^{0} + {3}^{1} + {3}^{2} + ... + {3}^{42} = \\ = \frac{1\cdot ( {3}^{43} - 1)}{3 - 1} = \frac{{3}^{43} - 1}{2}[/tex]
d)
[tex]b_{1} = 8, q = 8, n = 88[/tex]
[tex]d = 8 + {8}^{2} + {8}^{3} + ... + {8}^{88} = \\ = \frac{8\cdot ( {8}^{88} - 1)}{8 - 1} = \frac{8\cdot ( {8}^{88} - 1)}{7}[/tex]
