Question

47 Calculați a) (-2)¹+(-2)²+(-2)³+...+(-2)¹⁰⁰=?
b) (-3)¹+(-3)²+(-3)³+...+(-3)⁷⁵=?​

Answer 1

Răspuns:

47

Explicație pas cu pas:

a)

${( - 2)}^{1} + {( - 2)}^{2} + {( - 2)}^{3} + ... + {( - 2)}^{100} =$

$= - {2}^{1} + {2}^{2} - {2}^{3} + {2}^{4} - {2}^{5} + {2}^{6} - ... - {2}^{97} + {2}^{98} - {2}^{99} + {2}^{100}$

$= - ({2}^{1} + {2}^{3} + {2}^{5} + ... + {2}^{97} + {2}^{99}) + ({2}^{2} + {2}^{4} + {2}^{6} + ... + {2}^{98} + {2}^{100} )$

$= - 2 \cdot \underbrace{(1 + {2}^{2} + {2}^{4} + ... + {2}^{96} + {2}^{98})}_{S} + {2}^{2}\cdot \underbrace{(1 + {2}^{2} + {2}^{4} + ... + {2}^{96} + {2}^{98})}_{S}$

$= - 2S + 4S = 2S$

$S = 1 + {2}^{2} + {2}^{4} + ... + {2}^{96} + {2}^{98} = \\ = 1 + {4}^{1} + {4}^{2} + ... + {4}^{48} + {4}^{49}$

$4S = {4}^{1} + {4}^{2} + ... + {4}^{49} + {4}^{50}$

$4S + 1 = 1 + {4}^{1} + {4}^{2} + ... + {4}^{49} + {4}^{50}$

$4S + 1 = S + {4}^{50}$

$3S = {4}^{50} - 1 \implies S = \frac{{4}^{50} - 1}{3} \\ \implies \bf a = \frac{2({4}^{50} - 1)}{3} \iff a = \frac{2({2}^{100} - 1)}{3}$

sau:

progresie geometrică

$b_{1} = -2; q = -2; n = 100$

$a = {( - 2)}^{1} + {( - 2)}^{2} + {( - 2)}^{3} + ... + {( - 2)}^{100} = \\ = \frac{( - 2) \cdot \Big[{( - 2)}^{100} - 1\Big]}{( - 2) - 1} = \frac{{( - 2)}^{101} + 2}{ - 3} \\ = \frac{ - {2}^{101} + 2}{ - 3} = \bf \frac{{2}^{101} - 2}{3}$

b)

$b = {( - 3)}^{1} + {( - 3)}^{2} + {( - 3)}^{3} + ... + {( - 3)}^{75} =$

$= - {3}^{1} + {3}^{2} - {3}^{3} + {3}^{4} - {3}^{5} + {3}^{6} - ... - {3}^{71} + {3}^{72} - {3}^{73} + {3}^{74} - {3}^{75} = - ({3}^{1} + {3}^{3} + {3}^{5} + ... + {3}^{71} + {3}^{73}) + ({3}^{2} + {3}^{4} + {3}^{6} + ... + {3}^{72} + {3}^{74} ) - {3}^{75}$

$= - 3(1 + {3}^{2} + {3}^{4} + ... + {3}^{70} + {3}^{72}) + {3}^{2}(1 + {3}^{2} + {3}^{4} + ... + {3}^{70} + {3}^{72}) - {3}^{75}$

$= - 3S + 9S - {3}^{75} = 6S - {3}^{75}$

$S = 1 + {3}^{2} + {3}^{4} + ... + {3}^{70} + {3}^{72} = \\ = 1 + {9}^{1} + {9}^{2} + ... + {9}^{35} + {9}^{36}$

$9S = {9}^{1} + {9}^{2} + ... + {9}^{35} + {9}^{36} + {9}^{37}$

$9S + 1 = 1 + {9}^{1} + {9}^{2} + ... + {9}^{35} + {9}^{36} + {9}^{37}$

$9S + 1 = S + {9}^{37} \iff 8S = {9}^{37} - 1 \\ S = \frac{{9}^{37} - 1}{8} \iff S = \frac{{3}^{74} - 1}{8}$

$b = \frac{6({3}^{74} - 1)}{8} - {3}^{75} = \frac{3({3}^{74} - 1)}{4} - {3}^{75} = \frac{{3}^{75} - 3 - 4 \cdot {3}^{75}}{4} = \frac{ - 3 \cdot {3}^{75} - 3}{4}$

$\implies \bf b = - \frac{3({3}^{75} + 1)}{4}$