Solve It

$\quad \quad S=x+2{ x }^{ 2 }+{ 3x }^{ 3 }+...........+{ nx }^{ n }\quad \quad \quad \quad \quad \quad \quad \quad \qquad \quad \quad .....Eq.1\\ \quad xS=\quad \quad \quad { x }^{ 2 }+2{ x }^{ 3 }+{ 3x }^{ 4 }......+({ n-1)x }^{ n }+{ nx }^{ n+1 }\quad \quad \qquad .....Eq.2\\ Eq.1\quad -\quad Eq.2\\ (1-x)S=x+{ x }^{ 2 }+{ x }^{ 3 }+........+{ x }^{ n }-{ nx }^{ n+1 }\\ (1-x)S=\frac { x({ x }^{ n }-1) }{ x-1 } -{ nx }^{ n+1 }\\ Putting\quad x=2,\quad n=2014\\ -S=2({ 2 }^{ 2014 }-1)-2014({ 2 }^{ 2015 })\\ -S={ 2 }^{ 2015 }-2-2014({ 2 }^{ 2015 })\\ \quad S=2+2014({ 2 }^{ 2015 })-{ 2 }^{ 2015 }\\ \quad S=2+2013({ 2 }^{ 2015 })\\ so,\quad a=2,c=2013,d=2015\\ thus,\quad \frac { c+d }{ a } =\frac { 2013+2015 }{ 2 } =2014$

x + 2X^2 +3x^3 +.............................+nx^n=

x( 1+2x +3x^2+..................................+nx^(n-1) )+=

x(x+x^2 +x^3 +.........................................x^n)'=

x((x^(n+1) - x)/(x-1))' =

x((1 - (n+1)x^n + nx^(n+1))/(x-1)^2)

solving for x=2 and n=2014

we get

S= 2 + 2013*2^2015

so

a=2,c=2013,d=2015

so (b+c)/a = 2014