*** QuickLaTeX cannot compile formula:
\[
\begin{array}{l}
P(x) = (x^2 + x + 1)^{2015} + x + 1 \\
Calculati]
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Emergency stop.
(X^2+X+1)^2015+X+1=∑_(K=1)^2015▒(C_(2015.)^K X^(2.(2015-K) ).(X+1)^k) +X+1
=C_2015^0.X^4030+C_2015^1.X^4028.(X+1)+ C_2015^2.X^4026(X+1)^2+………..+ C_2015^2014,X^2(X+1)^2014+
C_2015^2015.(X+1)^2015+X+1=X^4030+2015.X^4029+(2015+C_2015^2).X^4028+…………….+2015(
∑_(k=0)^2014▒( C_2014^k,x^2.X^(4028-k))+(x+1)^2015+x+1=x^4030+2015x^4029+……+2016.x+2=0 fie
Cele 4030radacini ale ec.;x1,x2…………………..x4030. Impartim ec cu x^4030si notamy=1/xNoua ec va fi ;
2y^4030+2016yˆ^4029+…………………..=0si∑_(k=1)^4030▒〖yk=-2016/2=-1008〗unde yk=1/xk
Multumesc!
Mai simplu, avem
iar suma cerută, dacă aducem la acelasi numitor, e raportul ultimelor două sume Viete, deci