Question

Sa se scrie 2017^2 ca suma a 2017 nr consecutive

Answer 1

Răspuns:

$x \: . \: \: x + 1. \: \: x + 2. \: ... \: \: x + 2016$

${2017}^{2017} = x + x + 1 + x + 2 + ... + x + 2016 = > {2017}^{2017} = 2017x + 2017$

× 2016/2

$= > x = {2017}^{2016} - 1008$

$x + 1 = {2017}^{2016}$

$x + 2 = {2017}^{2016} - 1006$

$...$

$x + 2016 = {2017}^{2016} + 1008$

${2017}^{2017} = ( {2017}^{2016} - 1008 ) + ( {2017}^{2016} - 1007) + ... + {2017}^{2016} + ( {2017}^{2016} + 1) + ... + ( {2017}^{2016} + 1008)$

Mult succes în continuare!

Answer 2

$\it \dfrac{2017-1}{2}=\dfrac{2016}{2}=1008$

Suma celor 2017 numere consecutive este:

$\it (2017-1008)+(2017-1007)+(2017-1006)+\ ...\ +(2017-1)+2017+\\ \\ +(2017+1)+(2017+2)+\ ...\ +(2017+1007)+(2017+1008)$

După eliminarea parantezelor și reducerea termenilor opuși, obținem:

$\it \underbrace{\it2017+2017+2017+\ ...\ +2017}_{2017\ termeni} =2017\cdot2017=2017^2$