Question

1+3+3·4+3·4^2+3·4^3+...+3·4^2017 ​

Answer 1

Explicație pas cu pas:

progresie geometrică:

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

$b_{1} = 1; q = 4; n = 2018$

=>

$1 + 3 + 3 \cdot 4 + 3 \cdot {4}^{2} + 3 \cdot {4}^{3} + ... + 3 \cdot {4}^{2017} = 1 + 3 \cdot ( {4}^{0} + {4}^{1} + {4}^{2} + {4}^{3} + ... + {4}^{2017}) = 1 + 3 \cdot \frac{{4}^{2018} - 1}{4 - 1} = 1 + 3 \cdot \frac{{4}^{2018} - 1}{3} = 1 + {4}^{2018} - 1 = \bf {4}^{2018}$

Answer 2

Explicație pas cu pas:

(1+3)+3•4+3•4^2+3•4^3+...+3•4^2017= (4+3•4)+3•4^2+3•4^3+...+3•4^2017= (4^2+3•4^2)+...3•4^2017=... =4^2017+3•4^2017=4^2018