Question

Se considera expresia E(x)=2(x+3)²-3(x-1)(x+3)+(x-2)²-31 unde x este numar real. Calculati valoarea absoluta a numarului A= E(1)-E(2)+E(3)-E(4).....+E(2019)-E(2020)

Answer 1

Answer:

The answer is "5051".

Step-by-step explanation:

Given equation:

$\bold{E(x)= 2(x+3)^2-3(x-1)(x+3)+(x-2)^2-31}$

put the 1, 2, 3, ........2019, 20202:

When E(1),

$E(1) =2(1+3)^2-3(1-1)(1+3)+(1-2)^2-31$

        $=2(4)^2-3(0)(4)+(-1)^2-31\\\\=2(16)-0+1-31\\\\=32+1-31\\\\=2$

When E(2),

$E(2) =2(2+3)^2-3(2-1)(2+3)+(2-2)^2-31$

        $=2(5)^2-3(1)(5)+(0)^2-31\\\\=2(25)-15+0-31\\\\=50-15-31\\\\=4$

When E(3),

$E(3) =2(3+3)^2-3(3-1)(3+3)+(3-2)^2-31$

        $=2(6)^2-3(2)(6)+(1)^2-31\\\\=2(36)-36+1-31\\\\=72-66\\\\=6$

When E(2019),

$E(2019) =2(2019+3)^2-3(2019-1)(2019+3)+(2019-2)^2-31$

        $=2(2022)^2-3(2018)(2022)+(2017)^2-31\\\\=8,176,968-12,241,188+4,068,289-31 \\\\= 4038$

When E(2020),

$E(2020) =2(2020+3)^2-3(2020-1)(2020+3)+(2020-2)^2-31$

             $=2(2023)^2-3(2019)(2023)+(2018)^2-31\\\\=8,185,058-12,253,311+4,072,324-31\\\\=4040$

$\ Calculate: \\\\ \bold{A= E(1)-E(2)+E(3)-E(4).....+E(2019)-E(2020)}\\\\A= 2-4+6-8.....+4038-4040$

$A= (2+6....+4038)-(4+ 8+12...4040)$

Formula:

$t_n= a+(n-1)d\\\\\ When \\ \\ T_n =4040 \\ a= 4 \\ d= 4\\4040= 4+(n-1)4\\4040=4+4n-4\\4n=4040\\n= \frac{4040}{4}\\n= 1010\\\\When \\\\T_n =4038 \\ a= 2 \\ d= 4\\4038= 2+(n-1)4\\4038=2+4n-4\\4038=-2+4n\\4036=4n \\ n= \frac{4036}{4}\\n= 1009$

Formula for sum:

$When, n= 1010\\ a=4 \\ l=4040\\ \\ S= \frac{n}{2}(2a+l)\\\\S= \frac{1010}{2} (2\times 4+4040)\\\\S= 505(8+4040)\\\\S_1= 2044240$

$When, n= 1009\\ a=2 \\ l=4038\\ \\ S= \frac{n}{2}(2a+l)\\\\S= \frac{1009}{2} (2\times 2+4038)\\\\S= 504.5(4+4038)\\\\S_2= 2039189$

$S= S_1-S_2\\\\S= 2044240 -2039189\\\\S= 5051$

The absolute value is = 5051