Question

2. Se consideră expresia E(x) = (2x-1)²-(2-x)(2x+5)-(x+3)² + x(9-4x) +18, unde xe R. D

a) Arată că E(x)=x², pentru orice număr real x

b) Calculează suma S=E(2⁰) +E(2¹)+E(2²) + E(2³) + E(2⁴)+...+E(2⁵⁰).​

Answer 1

Explicație pas cu pas:

a)

$E(x) = (2x-1)^{2} - (2 - x)(2x + 5) - (x + 3)^{2} + x(9 - 4x) + 18$

$= 4 {x}^{2} - 4x + 1 - 4x - 10 + 2 {x}^{2} + 5x - {x}^{2} - 6x - 9 + 9x - 4 {x}^{2} + 18$

$= 6{x}^{2} - 5 {x}^{2} + 14x - 14x + 19 - 19$

$= {x}^{2}$

b)

$S = E(2⁰) + E(2¹) + E(2²) + E(2³) + E(2⁴) + ... + E(2⁵⁰)$

$= {( {2}^{0} )}^{2} + {({2}^{1})}^{2} + {( {2}^{2} )}^{2} + {( {2}^{3} )}^{2} + {( {2}^{4} )}^{2} + ... + {( {2}^{50} )}^{2}$

$= {2}^{0} + {2}^{2} + {2}^{4} + {2}^{6} + {2}^{8} + ... + {2}^{100}$

$= \frac{ {2}^{102} - 1}{3}$

cunoaștem formula:

$S_{n} = {2}^{0} + {2}^{1} + {2}^{2} + {2}^{3} + ... + {2}^{n} = {2}^{n + 1} - 1$

→

$S_{100} = {2}^{0} + {2}^{1} + {2}^{2} + {2}^{3} + ... + {2}^{100} = {2}^{101} - 1$

→

$S = S_{100} - ( {2}^{1} + {2}^{3} + {2}^{5} + ... + {2}^{101}) - {2}^{101}$

$S = S_{100} - 2({2}^{0} + {2}^{1} + {2}^{2} + {2}^{3} + ... + {2}^{100}) + {2}^{101}$

$= > S = S_{100} - 2S + {2}^{101}$

$3S = S_{100} + {2}^{101} < = > 3S = {2}^{101} - 1 + {2}^{101}$

$3S = 2 \times {2}^{101} - 1 < = > 3S = {2}^{102} - 1 \\ = > S = \frac{{2}^{102} - 1}{3}$