1-(2-(3-(...(2015-(2016-x)))))=2017. Created by Ангелина1802. algebra-ru

1-(2-(3-(...(2015-(2016-x)))))=2017...

100275

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1-(2-(3-(...(2015-(2016-x)))))=2017
1-2+(3-(...(2015-(2016-x))))=2017
1-2+3-...(2015-(2016-x)))=2017
...
1-2+3-...+2015-2016+x=2017
1+3+...+2015+x=2017+(2+4+...+2016)
уменьшаем 2, 4, ..., 2016 на 1, 3, ..., 2015
x=2017+(1+1+...+1) / единиц 2016/2=1008 штук
х=2017+1008=3025

2 votes Thanks 1

ShirokovP

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Через арифметическую прогрессию

Имеем две суммы
a1 = 1; a1008 = 2015
a1 = - 2 ; a1008 = - 2016

S1 = (a1 + a1008)/2*1008 = 504*(1 + 2015) = 2016*504 = 1016064
S2 = (a1 + a1008)/2*1008 = 504*( - 2 - 2016) = - 2018*504 = - 1017072
S = S1 + S2 = - 1008

x - 1008 = 2017
x = 2017 + 1008
x = 3025

3 votes Thanks 0

Ангелина1802 откуда вы взяли а1008 ?

ShirokovP a1 = 1; an = 2015; d = 2

ShirokovP an = a1 + d(n - 1) = 2015

ShirokovP 1 + 2(n - 1) = 2015

ShirokovP 1 + 2n - 2 = 2015

ShirokovP 2n - 1 = 2015

ShirokovP 2n = 2016

ShirokovP n = 1008

Ангелина1802 спасибо)

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