Question

(x-2)/2+(x-2)/6+(x-2)/12+...+(x-2)/(2014*2015)=2014

x=?

Answer 1

$\dfrac{x-2}{2}+\dfrac{x-2}{6}+\dfrac{x-2}{12}+...+\dfrac{x-2}{2014\cdot 2015} = 2014 \\ \\ \displaystyle \sum\limits_{k=1}^{2014}\dfrac{x-2}{k(k+1)} = 2014 \\ \\ (x-2)\sum\limits_{k=1}^{2014}\dfrac{1}{k(k+1)} = 2014\\ \\ \sum\limits_{k=1}^{2014}\dfrac{1}{k(k+1)} = \dfrac{2014}{x-2}\\ \\ \sum\limits_{k=1}^{2014}\Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big) = \dfrac{2014}{x-2}$

$\\ \dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{2014}-\dfrac{1}{2015} = \dfrac{2014}{x-2} \\ \\ 1 - \dfrac{1}{2015} = \dfrac{2014}{x-2} \\ \\ \dfrac{2014}{2015} = \dfrac{2014}{x-2} \\ \\ x-2 = 2015 \\ \\ \Rightarrow \boxed{x = 2017}$

Answer 2

Explicație pas cu pas:

(x-2)/2+(x-2)/6+(x-2)/12+.. +(x-2)/(2014×2015) = 2014

dăm factor comun pe (x-2)

(x-2)(1/2+1/6+1/12+...+1/(2014×2015)=2014

(x-2)(1/1×2 + 1/2×3 + 1/3×4 + ...+1/2014×2015) = 2014

(x-2)[(2-1)/1×2 + (3-2)/2×3 + (4-3)/3×4 +...+ (2015-2014)/2014×2015 = 2014

(x-2)(2/1×2 - 1/1×2 + 3/2×3 - 2/2×3 + 4/3×4 - 3/3×4 + ...+ 2015/2014×2015 - 2014/2014×2015) = 2014

se fac simplificarile și rămane:

(x-2)(1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/2014 - 1/2015) = 2014

se reduc fractiile asemenea cu + și -

(x-2)(1/1 - 1/2015) = 2014

(x-2)[(2015-1)/2015] = 2014

(x-2) × 2014/2015 = 2014 | : 2014

(x-2)/2015 = 1

x-2 = 2015

x = 2015 + 2

x = 2017