Răspuns:
[tex]\boldsymbol{ \red{S = \{1\}}}[/tex]
Explicație pas cu pas:
[tex]\dfrac{x+1}{2} + \dfrac{4x+2}{3} + \dfrac{9x+3}{4} + \dfrac{16x+4}{5} + ... +\dfrac{2012 \cdot 2012x+2012}{2013} = 1006 \cdot 2013[/tex]
Deoarece
[tex]1+2+...+2012 = \dfrac{2012 \cdot 2013}{2} = 1006 \cdot 2013\\[/tex]
Vom scrie:
[tex]\dfrac{x+1}{2} + \dfrac{4x+2}{3} + \dfrac{9x+3}{4} + ... +\dfrac{2012 \cdot 2012x+2012}{2013} = 1+2+3+...+2012[/tex]
Scădem termenii astfel:
[tex]\dfrac{x+1}{2} - 1 + \dfrac{4x+2}{3} - 2 + \dfrac{9x+3}{4} - 3 + ... +\dfrac{2012 \cdot 2012x+2012}{2013} - 2012 = 0[/tex]
[tex]\dfrac{x + 1 - 2}{2} + \dfrac{4x + 2 - 6}{3} + \dfrac{9x + 3 - 12}{4} + ... +\dfrac{2012^2 \cdot x + 2012 - 2012 \cdot 2013}{2013} = 0[/tex][tex]2012 - 2012 \cdot 2013 = 2012 (1 - 2013) = 2012 \cdot (-2012) = - 2012^2[/tex]
[tex]\dfrac{x - 1}{2} + \dfrac{4x - 4}{3} + \dfrac{9x - 9}{4} + ... +\dfrac{2012^2 \cdot x - 2012^2}{2013} = 0[/tex]
Dăm factor comun pe x - 1:
[tex]\dfrac{x - 1}{2} + \dfrac{4(x - 1)}{3} + \dfrac{9(x - 1)}{4} + ... +\dfrac{2012^2(x - 1)}{2013} = 0[/tex]
[tex](x-1) \cdot \bigg(\underbrace{\dfrac{1}{2} + \dfrac{4}{3} + \dfrac{9}{4} + ... +\dfrac{2012^2}{2013}}_{ > 0}\bigg) = 0[/tex]
[tex]\implies x - 1 = 0 \implies \bf x = 1[/tex]
Formule pentru diverse sume https://brainly.ro/tema/10771665
