Question

A= (2+4+6+8+...+4034) x ( 1÷1x2+ 1÷2x3 +...+1÷2017x2018) ÷ = supra COROANA

Answer 1

Atașez poza cu răspunsul :) A=2017^2

Answer 2

A= (2+4+6+8+...+4034) x (1÷1x2+ 1÷2x3 +...+1÷2017x2018)
S = 2+4+6+8+...+4034
nr de termeni=(MAX-min):ratie+1=(4034-2):2+1=4032:2+1= 2016+1=2017

S = 2+4+6+8+...+4034  (1)
S= 4034+4032+...+2 (2)
adunam cele doua relatii (1) cu (2) si obtinem:
2S=4036+4036+...+4036 (acest 4036 apare de 2017 ori, cati termeni avem);
deci 2S=4036*2017
impartim la 2 relatia si obtinem  S= 2018*2017;

M= 1÷1x2+ 1÷2x3 +...+1÷2017x2018 
aici se aplica formula generala: 1/[n*(n+1)]=1/n - 1/(n+1)

deci M=1/1-1/2 + 1/2-1/3+....+1/2017-1/2018  se reduc doi cate doi termeni si obtinem: M= 1/1 -1/2018= 2018/2018 - 1/2018= 2017/2018

Deci A = S * M= 2017*2018 * 2017/2018 = 2017*2017=2017^2

:D