Question

D=8+8 la a 2 a+8 la a 3 a+…+8 la 88.

Answer 1

Explicație pas cu pas:

progresie geometrică:

$S_{n} = \dfrac{b_{1}\cdot ({q}^{n} - 1) }{q - 1}$

$b_{1} = 8 \ , \ q = 8 \ , \ n = 88$

$D = 8 + {8}^{2} + {8}^{3} + ... + {8}^{88} = \dfrac{8\cdot ({8}^{88} - 1) }{8 - 1} = \dfrac{8\cdot ({8}^{88} - 1) }{7}$

sau:

$D = 8 + {8}^{2} + {8}^{3} + ... + {8}^{88}$

$8D = 8 \cdot (8 + {8}^{2} + {8}^{3} + ... + {8}^{88})$

$8D = {8}^{2} + {8}^{3} + {8}^{4} + ... + {8}^{88} + {8}^{89}$

$8D + 8 = \underbrace{8 + {8}^{2} + {8}^{3} + {8}^{4} + ... + {8}^{88}}_{D} + {8}^{89}$

$8D + 8 = D + {8}^{89} \iff 7D = {8}^{89} - 8 \\ \implies D = \frac{8({8}^{88} - 1)}{7}$