Question

Fie functia f:r→r, f(x)=x+1
Aratati ca numarul A=2012+2•[f(0)+f(1)+...f(2010)]este patrat perfect
Ajutorrr

Answer 1

[tex]f:R-\ \textgreater \ R ,f(x)=x+1\\ A=2012+2[f(0)+f(1)+f(2)+...+f(2010)]\\ A=2012+2[1+2+3+....+2011]\\ A=2012+2*\frac{2011*2012}{2}\\ A=2012+2011*2012\\ A=2012(1+2011) \Rightarrow A=2012^2[/tex]