Question

7.√2x + 1 unde x = 1+3+3²+...+ 3 ⁹⁹
va rog ajutor​

Answer 1

Explicație pas cu pas:

progresie geometrică:

$b_{1} = 1; q = 3; n = 100$

$x = 1 + {3}^{1} + {3}^{2} + {3}^{3} + ... + {3}^{99} = \frac{1 \cdot \Big({3}^{100} - 1\Big)}{3 - 1} = \frac{{3}^{100} - 1}{2}$

$\sqrt{2x + 1} = \sqrt{2 \cdot \frac{{3}^{100} - 1}{2} + 1} = \sqrt{{3}^{100} - 1 + 1} = \sqrt{ {3}^{100} } = \bf {3}^{50}$

sau:

$x = 1 + {3}^{1} + {3}^{2} + {3}^{3} + ... + {3}^{99}$

$3x = 3 \cdot (1 + {3}^{1} + {3}^{2} + ... + {3}^{99})$

$3x = {3}^{1} + {3}^{2} + {3}^{3} + ... + {3}^{99} + {3}^{100}$

$3x + 1 = \underbrace{1 + {3}^{1} + {3}^{2} + ... + {3}^{99}}_{x} + {3}^{100}$

$3x + 1 = x + {3}^{100} \iff 2x = {3}^{100} - 1 \\ \implies x = \frac{{3}^{100} - 1}{2}$

Answer 2

Răspuns:

=128 inta e răspunsul cos