Question

Ma poate ajuta cineva cu acest exercitiu?

Dau coroana!!

Answer 1

Explicație pas cu pas:

progresie geometrică

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

a)

$\sqrt{1 \cdot 7 + 6 \cdot 7 + 6 \cdot {7}^{2} + 6 \cdot {7}^{3} + ... + 6 \cdot {7}^{2019} } = {7}^{x}$

$\sqrt{1 \cdot 7 + 6 \cdot (7 + {7}^{2} + {7}^{3} + ... + {7}^{2019}) } = {7}^{x}$

$\sqrt{7 + 6 \cdot \dfrac{7 \cdot ({7}^{2019} - 1)}{7 - 1}} = {7}^{x}$

$\sqrt{7 + 6 \cdot \dfrac{{7}^{2020} - 7}{6}} = {7}^{x}$

$\sqrt{7 + {7}^{2020} - 7} = {7}^{x} \iff \sqrt{{7}^{1010\cdot 2} } = {7}^{x}$

${7}^{1010} = {7}^{x} \implies \bf x = 1010$

b)

$\sqrt{1 + 2 + 2 \cdot 3+ 2 \cdot {3}^{2} + 2 \cdot {3}^{3} + ... + 2 \cdot {3}^{2019} } = {3}^{x} \\ \sqrt{1 + 2 \cdot ({3}^{0} + {3}^{1} + {3}^{2} + {3}^{3} + ... + {3}^{2019}) } = {3}^{x} \\ \sqrt{1 + 2 \cdot \dfrac{1 \cdot ({3}^{2020} - 1)}{3 - 1}} = {3}^{x} \iff \sqrt{1 + \not2 \cdot \dfrac{{3}^{2020} - 1}{ \not2}} = {3}^{x} \\ \sqrt{1 + {3}^{2020} - 1} = {3}^{x} \iff \sqrt{{3}^{1010\cdot 2} } = {3}^{x}\\{3}^{1010} = {3}^{x} \implies \bf x = 1010$